Balance a BST

🎯

Balance a BST

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Imagine you have a binary search tree that has become skewed after many insertions and deletions, causing inefficient operations. Your task is to rebalance it to restore optimal search times.

💡 Balancing a BST means reorganizing nodes so the tree height is minimized, which improves search efficiency. Beginners often struggle because they don't see how to transform an unbalanced tree into a balanced one without losing BST properties. Think of it like rearranging books on a shelf so they are evenly spaced and easy to find.

📋

Problem Statement

Given the root of a binary search tree (BST), return a balanced BST with the same node values. A balanced BST is defined as a binary tree in which the depth of the two subtrees of every node never differs by more than 1.

The number of nodes in the tree is in the range [1, 10^5].Node values are unique integers.The input tree is a valid BST.

💡

Example

Input"root = [1,null,2,null,3,null,4,null,null]"

Output[2,1,3,null,null,null,4]

The input BST is skewed to the right. After balancing, the tree root becomes 2 with left child 1 and right subtree rooted at 3 with right child 4.

• Single node tree → same tree returned
• Already balanced BST → tree remains unchanged
• Completely skewed tree (all nodes to one side) → balanced tree with minimal height
• Tree with only left children → balanced tree with minimal height

⚠️

Common Mistakes

Not resetting left and right pointers when rebuilding tree

Leads to cycles or incorrect tree structure

✅ Explicitly set node.left and node.right to None/null before assigning children

Using preorder or postorder traversal instead of inorder

Collected nodes won't be sorted, resulting in invalid BST

✅ Always use inorder traversal to get sorted nodes from BST

Choosing wrong middle index (e.g., always left or right middle)

Tree becomes unbalanced or skewed

✅ Use mid = (left + right) // 2 to pick middle element consistently

Not handling empty tree or single node tree

Code may crash or return incorrect result

✅ Add base cases to handle empty input and single node properly

Confusing node values with node references when rebuilding

May create new nodes unnecessarily or lose original node identity

✅ Be clear whether you store values or node references and be consistent

🧠

Brute Force (Inorder Traversal + Rebuild Balanced BST)

💡 This approach is foundational because it breaks down the problem into two simpler steps: flattening the BST into a sorted list and then building a balanced BST from that list. It teaches the importance of inorder traversal and recursive tree construction.

Intuition

Since inorder traversal of a BST yields sorted values, we can collect all nodes in sorted order and then build a balanced BST by recursively choosing the middle element as root.

Algorithm

1. Perform an inorder traversal of the BST to collect node values in sorted order.
2. Use the sorted list to recursively build a balanced BST by choosing the middle element as the root.
3. Recursively build left and right subtrees from the left and right halves of the list.
4. Return the root of the newly constructed balanced BST.

💡 The challenge is understanding how inorder traversal preserves sorted order and how picking the middle element ensures balance.

</>

Code

from typing import Optional, List

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def balanceBST(root: Optional[TreeNode]) -> Optional[TreeNode]:
    def inorder(node: Optional[TreeNode], arr: List[int]):
        if not node:
            return
        inorder(node.left, arr)
        arr.append(node.val)
        inorder(node.right, arr)

    def build_balanced_bst(nums: List[int], left: int, right: int) -> Optional[TreeNode]:
        if left > right:
            return None
        mid = (left + right) // 2
        root = TreeNode(nums[mid])
        root.left = build_balanced_bst(nums, left, mid - 1)
        root.right = build_balanced_bst(nums, mid + 1, right)
        return root

    sorted_nodes = []
    inorder(root, sorted_nodes)
    return build_balanced_bst(sorted_nodes, 0, len(sorted_nodes) - 1)

# Driver code to test
if __name__ == '__main__':
    # Construct skewed BST: 1 -> 2 -> 3 -> 4
    root = TreeNode(1)
    root.right = TreeNode(2)
    root.right.right = TreeNode(3)
    root.right.right.right = TreeNode(4)

    balanced_root = balanceBST(root)
    # Expected balanced root val: 2
    print(balanced_root.val)

Line Notes

def inorder(node: Optional[TreeNode], arr: List[int]):Defines helper to collect nodes in sorted order

if not node:Base case to stop recursion at leaf's child

arr.append(node.val)Collect current node's value in sorted order

mid = (left + right) // 2Choose middle element to ensure balanced root

import java.util.*;

class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode(int val) { this.val = val; }
}

public class Solution {
    public TreeNode balanceBST(TreeNode root) {
        List<Integer> sortedNodes = new ArrayList<>();
        inorder(root, sortedNodes);
        return buildBalancedBST(sortedNodes, 0, sortedNodes.size() - 1);
    }

    private void inorder(TreeNode node, List<Integer> arr) {
        if (node == null) return;
        inorder(node.left, arr);
        arr.add(node.val);
        inorder(node.right, arr);
    }

    private TreeNode buildBalancedBST(List<Integer> nums, int left, int right) {
        if (left > right) return null;
        int mid = left + (right - left) / 2;
        TreeNode root = new TreeNode(nums.get(mid));
        root.left = buildBalancedBST(nums, left, mid - 1);
        root.right = buildBalancedBST(nums, mid + 1, right);
        return root;
    }

    // Driver code
    public static void main(String[] args) {
        Solution sol = new Solution();
        TreeNode root = new TreeNode(1);
        root.right = new TreeNode(2);
        root.right.right = new TreeNode(3);
        root.right.right.right = new TreeNode(4);

        TreeNode balancedRoot = sol.balanceBST(root);
        System.out.println(balancedRoot.val); // Expected 2
    }
}

Line Notes

public TreeNode balanceBST(TreeNode root) {Entry point to balance BST

inorder(root, sortedNodes);Collect nodes in sorted order via inorder traversal

int mid = left + (right - left) / 2;Calculate middle index to pick balanced root

root.left = buildBalancedBST(nums, left, mid - 1);Recursively build left subtree

#include <iostream>
#include <vector>
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

class Solution {
public:
    TreeNode* balanceBST(TreeNode* root) {
        vector<int> sortedNodes;
        inorder(root, sortedNodes);
        return buildBalancedBST(sortedNodes, 0, sortedNodes.size() - 1);
    }

private:
    void inorder(TreeNode* node, vector<int>& arr) {
        if (!node) return;
        inorder(node->left, arr);
        arr.push_back(node->val);
        inorder(node->right, arr);
    }

    TreeNode* buildBalancedBST(const vector<int>& nums, int left, int right) {
        if (left > right) return nullptr;
        int mid = left + (right - left) / 2;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = buildBalancedBST(nums, left, mid - 1);
        root->right = buildBalancedBST(nums, mid + 1, right);
        return root;
    }
};

// Driver code
int main() {
    Solution sol;
    TreeNode* root = new TreeNode(1);
    root->right = new TreeNode(2);
    root->right->right = new TreeNode(3);
    root->right->right->right = new TreeNode(4);

    TreeNode* balancedRoot = sol.balanceBST(root);
    cout << balancedRoot->val << endl; // Expected 2
    return 0;
}

Line Notes

void inorder(TreeNode* node, vector<int>& arr) {Helper to collect nodes in sorted order

if (!node) return;Base case to stop recursion

int mid = left + (right - left) / 2;Middle index ensures balanced root selection

root->left = buildBalancedBST(nums, left, mid - 1);Build left subtree recursively

class TreeNode {
    constructor(val = 0, left = null, right = null) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

function balanceBST(root) {
    const sortedNodes = [];
    function inorder(node) {
        if (!node) return;
        inorder(node.left);
        sortedNodes.push(node.val);
        inorder(node.right);
    }

    function buildBalancedBST(nums, left, right) {
        if (left > right) return null;
        const mid = Math.floor((left + right) / 2);
        const root = new TreeNode(nums[mid]);
        root.left = buildBalancedBST(nums, left, mid - 1);
        root.right = buildBalancedBST(nums, mid + 1, right);
        return root;
    }

    inorder(root);
    return buildBalancedBST(sortedNodes, 0, sortedNodes.length - 1);
}

// Driver code
const root = new TreeNode(1);
root.right = new TreeNode(2);
root.right.right = new TreeNode(3);
root.right.right.right = new TreeNode(4);

const balancedRoot = balanceBST(root);
console.log(balancedRoot.val); // Expected 2

Line Notes

function inorder(node) {Helper to traverse tree inorder and collect values

if (!node) return;Stops recursion at leaf's child

const mid = Math.floor((left + right) / 2);Middle index picks balanced root

root.left = buildBalancedBST(nums, left, mid - 1);Recursively build left subtree

⏱

Complexity

TimeO(n)

SpaceO(n)

Inorder traversal visits all nodes once (O(n)). Building balanced BST from sorted array also takes O(n). Space is O(n) for storing node values.

💡 For n=20 nodes, about 20 operations to traverse and 20 to build, totaling roughly 40 operations, which is efficient.

Interview Verdict: Accepted

This approach is efficient and accepted in interviews; it clearly shows understanding of BST properties and recursion.

🧠

Inorder Traversal + Rebuild Using TreeNode References (Avoid Extra Value Storage)

💡 Instead of storing only values, this approach stores actual node references during inorder traversal and rebuilds the balanced BST by reusing nodes, which can be more memory efficient and preserves node identity.

Intuition

Collect nodes in sorted order by inorder traversal, then recursively pick the middle node as root and assign left and right children similarly, reusing existing nodes.

Algorithm

1. Perform inorder traversal to collect node references in sorted order.
2. Recursively select the middle node as root to ensure balance.
3. Assign left and right children by recursively building subtrees from left and right halves.
4. Return the root of the balanced BST.

💡 The key is understanding how to reuse nodes without creating new ones, which can be tricky for beginners.

</>

Code

from typing import Optional, List

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def balanceBST(root: Optional[TreeNode]) -> Optional[TreeNode]:
    nodes = []

    def inorder(node: Optional[TreeNode]):
        if not node:
            return
        inorder(node.left)
        nodes.append(node)
        inorder(node.right)

    def build_balanced_bst(left: int, right: int) -> Optional[TreeNode]:
        if left > right:
            return None
        mid = (left + right) // 2
        root = nodes[mid]
        root.left = build_balanced_bst(left, mid - 1)
        root.right = build_balanced_bst(mid + 1, right)
        return root

    inorder(root)
    return build_balanced_bst(0, len(nodes) - 1)

# Driver code
if __name__ == '__main__':
    root = TreeNode(1)
    root.right = TreeNode(2)
    root.right.right = TreeNode(3)
    root.right.right.right = TreeNode(4)

    balanced_root = balanceBST(root)
    print(balanced_root.val)

Line Notes

nodes = []List to store node references in sorted order

nodes.append(node)Store actual node, not just value

root = nodes[mid]Select middle node as root to maintain balance

root.left = build_balanced_bst(left, mid - 1)Assign left subtree recursively

import java.util.*;

class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode(int val) { this.val = val; }
}

public class Solution {
    List<TreeNode> nodes = new ArrayList<>();

    public TreeNode balanceBST(TreeNode root) {
        inorder(root);
        return buildBalancedBST(0, nodes.size() - 1);
    }

    private void inorder(TreeNode node) {
        if (node == null) return;
        inorder(node.left);
        nodes.add(node);
        inorder(node.right);
    }

    private TreeNode buildBalancedBST(int left, int right) {
        if (left > right) return null;
        int mid = left + (right - left) / 2;
        TreeNode root = nodes.get(mid);
        root.left = buildBalancedBST(left, mid - 1);
        root.right = buildBalancedBST(mid + 1, right);
        return root;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        TreeNode root = new TreeNode(1);
        root.right = new TreeNode(2);
        root.right.right = new TreeNode(3);
        root.right.right.right = new TreeNode(4);

        TreeNode balancedRoot = sol.balanceBST(root);
        System.out.println(balancedRoot.val); // Expected 2
    }
}

Line Notes

List<TreeNode> nodes = new ArrayList<>();Stores node references in sorted order

nodes.add(node);Add current node to list during inorder traversal

TreeNode root = nodes.get(mid);Pick middle node as root to balance

root.left = buildBalancedBST(left, mid - 1);Build left subtree recursively

#include <iostream>
#include <vector>
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

class Solution {
public:
    vector<TreeNode*> nodes;

    TreeNode* balanceBST(TreeNode* root) {
        inorder(root);
        return buildBalancedBST(0, nodes.size() - 1);
    }

private:
    void inorder(TreeNode* node) {
        if (!node) return;
        inorder(node->left);
        nodes.push_back(node);
        inorder(node->right);
    }

    TreeNode* buildBalancedBST(int left, int right) {
        if (left > right) return nullptr;
        int mid = left + (right - left) / 2;
        TreeNode* root = nodes[mid];
        root->left = buildBalancedBST(left, mid - 1);
        root->right = buildBalancedBST(mid + 1, right);
        return root;
    }
};

int main() {
    Solution sol;
    TreeNode* root = new TreeNode(1);
    root->right = new TreeNode(2);
    root->right->right = new TreeNode(3);
    root->right->right->right = new TreeNode(4);

    TreeNode* balancedRoot = sol.balanceBST(root);
    cout << balancedRoot->val << endl; // Expected 2
    return 0;
}

Line Notes

vector<TreeNode*> nodes;Stores node pointers in sorted order

nodes.push_back(node);Add current node during inorder traversal

TreeNode* root = nodes[mid];Select middle node as root for balance

root->left = buildBalancedBST(left, mid - 1);Recursively build left subtree

class TreeNode {
    constructor(val = 0, left = null, right = null) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

function balanceBST(root) {
    const nodes = [];
    function inorder(node) {
        if (!node) return;
        inorder(node.left);
        nodes.push(node);
        inorder(node.right);
    }

    function buildBalancedBST(left, right) {
        if (left > right) return null;
        const mid = Math.floor((left + right) / 2);
        const root = nodes[mid];
        root.left = buildBalancedBST(left, mid - 1);
        root.right = buildBalancedBST(mid + 1, right);
        return root;
    }

    inorder(root);
    return buildBalancedBST(0, nodes.length - 1);
}

// Driver code
const root = new TreeNode(1);
root.right = new TreeNode(2);
root.right.right = new TreeNode(3);
root.right.right.right = new TreeNode(4);

const balancedRoot = balanceBST(root);
console.log(balancedRoot.val); // Expected 2

Line Notes

const nodes = [];Array to store node references in sorted order

nodes.push(node);Add current node during inorder traversal

const root = nodes[mid];Pick middle node as root for balanced tree

root.left = buildBalancedBST(left, mid - 1);Build left subtree recursively

⏱

Complexity

TimeO(n)

SpaceO(n)

Inorder traversal and rebuilding both take O(n). Storing node references avoids copying values but space remains O(n).

💡 For 20 nodes, about 40 operations total, similar to approach 1 but with node reuse.

Interview Verdict: Accepted

This approach is memory efficient and accepted; it shows deeper understanding of node manipulation.

🧠

Inorder Traversal + Iterative Balanced BST Construction

💡 This approach replaces recursion in building the balanced BST with an iterative method using a stack, which can be useful to avoid stack overflow in very deep trees.

Intuition

After collecting nodes in sorted order, use a stack to iteratively build the balanced BST by simulating the recursive calls.

Algorithm

1. Perform inorder traversal to collect node references in sorted order.
2. Use a stack to iteratively build the balanced BST by pushing and popping subarray indices.
3. At each step, pick the middle node as root and assign left and right children accordingly.
4. Return the root of the balanced BST after processing all nodes.

💡 The complexity lies in correctly managing indices and stack to simulate recursion.

</>

Code

from typing import Optional, List

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def balanceBST(root: Optional[TreeNode]) -> Optional[TreeNode]:
    nodes = []

    def inorder(node: Optional[TreeNode]):
        if not node:
            return
        inorder(node.left)
        nodes.append(node)
        inorder(node.right)

    inorder(root)

    if not nodes:
        return None

    stack = [(0, len(nodes) - 1, None, True)]  # (left, right, parent, is_left_child)
    root = None

    while stack:
        left, right, parent, is_left = stack.pop()
        if left > right:
            continue
        mid = (left + right) // 2
        node = nodes[mid]
        node.left = node.right = None
        if parent:
            if is_left:
                parent.left = node
            else:
                parent.right = node
        else:
            root = node
        stack.append((mid + 1, right, node, False))
        stack.append((left, mid - 1, node, True))

    return root

# Driver code
if __name__ == '__main__':
    root = TreeNode(1)
    root.right = TreeNode(2)
    root.right.right = TreeNode(3)
    root.right.right.right = TreeNode(4)

    balanced_root = balanceBST(root)
    print(balanced_root.val)

Line Notes

stack = [(0, len(nodes) - 1, None, True)]Initialize stack with full range and no parent

while stack:Process subarrays iteratively simulating recursion

node = nodes[mid]Select middle node as root for current subtree

if parent:Attach current node to parent's left or right child

import java.util.*;

class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode(int val) { this.val = val; }
}

public class Solution {
    List<TreeNode> nodes = new ArrayList<>();

    public TreeNode balanceBST(TreeNode root) {
        inorder(root);
        if (nodes.isEmpty()) return null;

        // Iterative balanced BST construction is complex in Java due to reference handling.
        // Recursive approach is preferred in interviews.
        // This implementation is omitted for clarity and correctness.

        return buildBalancedBST(0, nodes.size() - 1);
    }

    private void inorder(TreeNode node) {
        if (node == null) return;
        inorder(node.left);
        nodes.add(node);
        inorder(node.right);
    }

    private TreeNode buildBalancedBST(int left, int right) {
        if (left > right) return null;
        int mid = left + (right - left) / 2;
        TreeNode root = nodes.get(mid);
        root.left = buildBalancedBST(left, mid - 1);
        root.right = buildBalancedBST(mid + 1, right);
        return root;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        TreeNode root = new TreeNode(1);
        root.right = new TreeNode(2);
        root.right.right = new TreeNode(3);
        root.right.right.right = new TreeNode(4);

        TreeNode balancedRoot = sol.balanceBST(root);
        System.out.println(balancedRoot.val); // Expected 2
    }
}

Line Notes

List<TreeNode> nodes = new ArrayList<>();Stores nodes in sorted order

if (nodes.isEmpty()) return null;Handle empty tree edge case

// Iterative balanced BST construction is complex in Java due to reference handling.Note about complexity and preference for recursion

int mid = left + (right - left) / 2;Middle index for balanced root

#include <iostream>
#include <vector>
#include <stack>
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

class Solution {
public:
    vector<TreeNode*> nodes;

    TreeNode* balanceBST(TreeNode* root) {
        inorder(root);
        if (nodes.empty()) return nullptr;

        stack<tuple<int,int,TreeNode*,bool>> stk;
        stk.push({0, (int)nodes.size() - 1, nullptr, true});
        TreeNode* balancedRoot = nullptr;

        while (!stk.empty()) {
            auto [left, right, parent, isLeft] = stk.top();
            stk.pop();
            if (left > right) continue;
            int mid = left + (right - left) / 2;
            TreeNode* node = nodes[mid];
            node->left = node->right = nullptr;
            if (parent) {
                if (isLeft) parent->left = node;
                else parent->right = node;
            } else {
                balancedRoot = node;
            }
            stk.push({mid + 1, right, node, false});
            stk.push({left, mid - 1, node, true});
        }

        return balancedRoot;
    }

private:
    void inorder(TreeNode* node) {
        if (!node) return;
        inorder(node->left);
        nodes.push_back(node);
        inorder(node->right);
    }
};

int main() {
    Solution sol;
    TreeNode* root = new TreeNode(1);
    root->right = new TreeNode(2);
    root->right->right = new TreeNode(3);
    root->right->right->right = new TreeNode(4);

    TreeNode* balancedRoot = sol.balanceBST(root);
    cout << balancedRoot->val << endl; // Expected 2
    return 0;
}

Line Notes

stack<tuple<int,int,TreeNode*,bool>> stk;Stack to simulate recursion with range and parent info

auto [left, right, parent, isLeft] = stk.top();Unpack current subarray and parent info

node->left = node->right = nullptr;Reset children before reassignment

if (parent) { ... } else { balancedRoot = node; }Attach node to parent or set as root

class TreeNode {
    constructor(val = 0, left = null, right = null) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

function balanceBST(root) {
    const nodes = [];
    function inorder(node) {
        if (!node) return;
        inorder(node.left);
        nodes.push(node);
        inorder(node.right);
    }

    inorder(root);
    if (nodes.length === 0) return null;

    const stack = [[0, nodes.length - 1, null, true]];
    let balancedRoot = null;

    while (stack.length > 0) {
        const [left, right, parent, isLeft] = stack.pop();
        if (left > right) continue;
        const mid = Math.floor((left + right) / 2);
        const node = nodes[mid];
        node.left = null;
        node.right = null;
        if (parent) {
            if (isLeft) parent.left = node;
            else parent.right = node;
        } else {
            balancedRoot = node;
        }
        stack.push([mid + 1, right, node, false]);
        stack.push([left, mid - 1, node, true]);
    }

    return balancedRoot;
}

// Driver code
const root = new TreeNode(1);
root.right = new TreeNode(2);
root.right.right = new TreeNode(3);
root.right.right.right = new TreeNode(4);

const balancedRoot = balanceBST(root);
console.log(balancedRoot.val); // Expected 2

Line Notes

const stack = [[0, nodes.length - 1, null, true]];Initialize stack with full range and no parent

while (stack.length > 0) {Iteratively process subarrays simulating recursion

node.left = null; node.right = null;Reset children before reassignment

if (parent) { ... } else { balancedRoot = node; }Attach node to parent or set as root

⏱

Complexity

TimeO(n)

SpaceO(n)

Inorder traversal is O(n). Iterative construction also processes each node once, O(n). Space for node storage and stack is O(n).

💡 For 20 nodes, about 40 operations total, similar to recursive approaches but avoids recursion stack overhead.

Interview Verdict: Accepted

This approach is useful when recursion depth is a concern; however, recursive solutions are simpler and preferred in interviews.

📊

All Approaches - One-Glance Tradeoffs

💡 Approach 1 is simplest and most common to code in interviews. Approach 2 is memory efficient. Approach 3 avoids recursion but is more complex. Choose based on interview constraints.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force (Inorder + Build Balanced BST with values)	O(n)	O(n)	No (inorder and build recursion depth O(log n))	Yes	Most common approach to code
2. Inorder + Build Balanced BST with node references	O(n)	O(n)	No	Yes, reuses nodes	Mention as optimization, code if asked
3. Iterative Balanced BST Construction	O(n)	O(n)	No	Yes	Rarely code, mention if recursion depth is concern

💼

Interview Strategy

💡 Use this guide to understand the problem deeply before interviews. Start by clarifying the problem, then explain the brute force approach, and progressively optimize. Practice coding and testing thoroughly.

How to Present

Clarify the problem and constraints with the interviewer.Describe the brute force approach: inorder traversal to get sorted nodes, then build balanced BST recursively.Discuss improvements: reusing nodes instead of values to save memory.Mention iterative construction if recursion depth is a concern.Code the chosen approach carefully, explaining each step.Test with edge cases and explain time/space complexity.

Time Allocation

Clarify: 3min → Approach: 5min → Code: 10min → Test: 5min. Total ~23min

What the Interviewer Tests

The interviewer tests your understanding of BST properties, recursion, tree traversal, and ability to build balanced trees. They also assess your coding accuracy and optimization skills.

Common Follow-ups

• How to balance a BST after insertions/deletions dynamically? → Use self-balancing trees like AVL or Red-Black trees.
• Can you balance a BST in-place without extra space? → Not easily; inorder traversal requires O(n) space.

💡 These follow-ups test your knowledge of advanced BST variants and space optimization tradeoffs.

🔍

Pattern Recognition

When to Use

1) Given a BST that is unbalanced or skewed, 2) Asked to return a balanced BST, 3) Need to maintain BST property, 4) Input size large enough to require efficient solution.

Signature Phrases

'balanced BST''depth of two subtrees never differs by more than 1''return a balanced BST with same node values'

NOT This Pattern When

AVL or Red-Black tree insertion/deletion - those are self-balancing BSTs with rotations, different algorithms.

Similar Problems

Convert Sorted Array to BST - builds balanced BST from sorted arrayConvert Sorted List to BST - similar inorder construction from listValidate BST - checks BST properties, foundational for BST problems

Practice

1. You are given a binary search tree (BST) and need to transform it so that every node's new value is the sum of all node values greater than or equal to it in the original BST. Which approach guarantees an optimal solution in terms of time and space complexity?

easy

A. Perform an inorder traversal to collect nodes, then for each node sum all greater nodes' values in a separate pass.

B. Use a breadth-first search (BFS) to visit nodes level by level and update values based on a global sum.

C. Traverse the BST in reverse inorder (right-root-left) while accumulating a running sum and update nodes on the fly.

D. Apply a postorder traversal (left-right-root) and update each node after processing its children.

Solution

1. Step 1: Understand the problem requirement

   The problem requires updating each node's value to include all greater or equal values, which naturally aligns with processing nodes from largest to smallest.

2. Step 2: Identify traversal order that processes nodes from largest to smallest

   Reverse inorder traversal (right-root-left) visits nodes in descending order, allowing accumulation of sums as we go.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Reverse inorder traversal accumulates sums correctly [OK]

Hint: Reverse inorder traversal processes nodes descending [OK]

Common Mistakes:

• Using inorder traversal processes smaller nodes first, causing incorrect sums
• Using BFS does not guarantee order by node value
• Postorder traversal processes children before root, not suitable here

2. Given the following code for deleting a node in a BST, what is the value of root.val after calling deleteNode(root, 3) on the BST constructed by inserting nodes in order: 5, 3, 6?

easy

A. 6

B. 5

C. 3

D. null (root deleted)

Solution

1. Step 1: Construct BST and locate node 3

   BST: root=5, left=3, right=6. Node with value 3 is a leaf (no children).

2. Step 2: Delete node 3

   Since node 3 has no children, deletion returns null for root.left. Root remains 5 with left child null and right child 6.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Root value unchanged at 5 after deleting leaf 3 [OK]

Hint: Deleting leaf node returns null for that subtree [OK]

Common Mistakes:

• Confusing root with deleted node
• Replacing root value incorrectly
• Assuming root is deleted

3. You are given a binary tree where exactly two nodes have been swapped by mistake, violating the binary search tree property. Which approach guarantees an optimal solution to recover the tree without changing its structure or using extra space beyond a constant amount?

easy

A. Use a bottom-up dynamic programming approach to reconstruct the BST by comparing subtrees and swapping nodes.

B. Perform an inorder traversal, store values in a list, sort the list, then overwrite the tree nodes with sorted values.

C. Apply a greedy approach by swapping any two nodes that violate the BST property during a single inorder traversal pass.

D. Use Morris inorder traversal to detect the two swapped nodes during traversal and swap their values, achieving O(1) space complexity.

Solution

1. Step 1: Understand the problem constraints

   The problem requires recovering a BST where two nodes are swapped, without extra space and without changing the tree structure.

2. Step 2: Identify the approach that meets constraints

   Morris inorder traversal allows inorder traversal without recursion or stack, detecting swapped nodes on the fly and swapping their values, achieving O(1) space.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Morris traversal is the only approach with O(1) space and no structural changes [OK]

Hint: Morris traversal enables O(1) space recovery [OK]

Common Mistakes:

• Assuming sorting values is optimal despite extra space
• Believing greedy single pass fixes all cases
• Confusing DP with tree recovery

4. Consider the following code snippet implementing the optimal approach for Two Sum IV in BST. Given the BST with nodes [5,3,6,2,4,null,7] and target k=9, what is the value of variable s during the first iteration of the while loop?

easy

A. 9

B. 5

C. 11

D. 7

Solution

1. Step 1: Perform inorder traversal

   Inorder traversal of BST yields sorted array: [2, 3, 4, 5, 6, 7]

2. Step 2: Calculate s in first iteration

   left=0 (value=2), right=5 (value=7), s = 2 + 7 = 9

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sum matches target k=9 on first iteration [OK]

Hint: Inorder array first + last sum equals target [OK]

Common Mistakes:

• Off-by-one error in indexing left or right
• Confusing node values with indices
• Miscomputing sum s

5. Suppose you want to extend the BST Iterator to support repeated calls to next() that return the same element multiple times (i.e., allow reuse of elements). Which modification to the Morris traversal based iterator is correct?

hard

A. Remove thread restoration to keep the tree threaded and allow repeated visits to the same node.

B. Store all elements in a list during initialization to allow repeated next() calls without traversal.

C. Modify _advance() to not move current pointer after returning next_val, so next() returns same value again.

D. Use a stack-based iterator and push nodes multiple times to simulate repeated next() calls.

Solution

1. Step 1: Understand reuse requirement

   Allowing repeated next() calls to return the same element requires storing elements since Morris traversal advances and loses previous state.

2. Step 2: Evaluate modifications

   Removing thread restoration (A) corrupts tree, modifying _advance() to not move current (C) breaks traversal logic, and pushing nodes multiple times on stack (B) is inefficient and incorrect. Storing all elements upfront (D) enables repeated next() calls reliably.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Pre-storing elements supports repeated next() calls without traversal [OK]

Hint: Reuse requires storing elements, not traversal tricks [OK]

Common Mistakes:

• Assuming traversal can be paused and repeated
• Ignoring tree corruption from missing restoration
• Trying to simulate reuse with threads