Find Bottom Left Tree Value

📋

Problem

Imagine you are looking at a tree from the bottom left side and want to find the value of the node that would be visible there.

Given the root of a binary tree, return the value of the leftmost node in the last row of the tree.

The number of nodes in the tree is in the range [1, 10^5].Node values are integers and can be positive, negative, or zero.

Edge cases: Single node tree → output is the root valueTree with only left children → output is the bottom-most left childTree with only right children → output is the bottom-most right child (which is also leftmost at that level)

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IDE

def findBottomLeftValue(root: TreeNode) -> int:public int findBottomLeftValue(TreeNode root)int findBottomLeftValue(TreeNode* root)function findBottomLeftValue(root)

def findBottomLeftValue(root):
    # Write your solution here
    pass

class Solution {
    public int findBottomLeftValue(TreeNode root) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int findBottomLeftValue(TreeNode* root) {
    // Write your solution here
    return 0;
}

function findBottomLeftValue(root) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: root value instead of bottom leftNot traversing beyond root or not updating leftmost node at deeper levels.✅ Update leftmost value at the start of each BFS level, not just once at root.

Wrong: left child of root instead of bottom leftGreedy approach assuming bottom left is always left child of root or first left child found.✅ Use BFS to traverse all levels and update leftmost node at each level.

Wrong: rightmost node at last levelNot tracking leftmost node correctly, possibly updating with last node at level.✅ Record the first node at each level during BFS, not the last.

Wrong: incorrect node due to off-by-one depth errorUpdating leftmost node at wrong depth or not strictly greater depth.✅ Update leftmost only when current depth > max_depth.

Wrong: TLE or stack overflowUsing DFS with exponential complexity or inefficient recursion on large input.✅ Implement BFS with queue for O(n) time complexity.

✓

Test Cases

Focus on base cases like single node and skewed trees to solidify your solution.

Watch out for common pitfalls like greedy choices and depth off-by-one errors.

Ensure your solution is efficient enough to handle large inputs within time limits.

t1_01basic

Input{"root":[2,1,3]}

Expected1

⏱ Performance - must finish in 2000ms

The last row has nodes 1 and 3, the leftmost is 1.

💡 Try a level order traversal (BFS) to visit nodes level by level.

💡 Track the first node value at each level to find the leftmost node.

💡 Return the first node value of the last level visited.

Why it failed: Returned wrong value; likely did not track leftmost node at last level correctly. Fix by updating leftmost value only at start of each level during BFS.

✓ Correctly identified the leftmost node at the bottom level.

t1_02basic

Input{"root":[1,2,3,4,5,6,7]}

Expected4

⏱ Performance - must finish in 2000ms

Complete binary tree; last row nodes are 4,5,6,7; leftmost is 4.

💡 Use BFS to process nodes level by level.

💡 At each level, record the first node's value as candidate leftmost.

💡 Return the candidate after processing all levels.

Why it failed: Incorrect output due to not updating leftmost node at last level. Fix by resetting leftmost value at each new level in BFS.

✓ Correctly found leftmost node in the last row of a complete binary tree.

t2_01edge

Input{"root":[10]}

Expected10

⏱ Performance - must finish in 2000ms

Single node tree; bottom left is the root itself.

💡 Check if tree has only one node.

💡 The root is the leftmost and bottom node by default.

💡 Return root's value immediately if no children.

Why it failed: Failed single node case; likely assumes multiple levels. Fix by handling single node tree as base case.

✓ Correctly handled single node tree.

t2_02edge

Input{"root":[1,2,null,3,null,4]}

Expected4

⏱ Performance - must finish in 2000ms

Tree with only left children; bottom-most left child is 4.

💡 Traverse left children deeply to find bottom left.

💡 BFS or DFS can be used; ensure left children are prioritized.

💡 Return value of the deepest leftmost node.

Why it failed: Incorrect output due to ignoring left subtree depth or order. Fix by prioritizing left children in traversal.

✓ Correctly found bottom-most left child in left-skewed tree.

t2_03edge

Input{"root":[1,null,2,null,3,null,4]}

Expected4

⏱ Performance - must finish in 2000ms

Tree with only right children; bottom-most right child is 4, which is also leftmost at that level.

💡 Even if only right children exist, last level's leftmost is that node.

💡 BFS visits nodes level by level; last level has one node.

💡 Return the value of the last node visited at the bottom level.

Why it failed: Wrong output due to assuming left child must exist. Fix by handling levels with single right child correctly.

✓ Correctly handled right-skewed tree.

t3_01corner

Input{"root":[1,2,3,4,null,null,5,6]}

Expected6

⏱ Performance - must finish in 2000ms

Greedy trap: last level nodes are 6 and 5; leftmost is 6 (left child of 4).

💡 Do not assume leftmost node is always left child of root's left subtree.

💡 Use BFS to explore all nodes level by level.

💡 Track first node at each level, not just left subtree.

Why it failed: Greedy approach failed by picking 4 or 5 instead of 6. Fix by using BFS level order traversal to find bottom left.

✓ Correctly avoided greedy trap by BFS traversal.

t3_02corner

Input{"root":[1,2,3,4,5,6,7,null,null,8]}

Expected8

⏱ Performance - must finish in 2000ms

Off-by-one depth error: bottom left is 8 at last level, not 4 or 6.

💡 Ensure depth counting includes all levels correctly.

💡 Track depth and update leftmost only when depth increases.

💡 Return leftmost node at maximum depth.

Why it failed: Off-by-one error in depth tracking caused wrong node selection. Fix by updating leftmost only when depth strictly increases.

✓ Correctly handled depth and leftmost node update.

t3_03corner

Input{"root":[1,2,3,4,null,null,5,6,null,null,7]}

Expected6

⏱ Performance - must finish in 2000ms

Confusion between 0/1 and unbounded: only one instance of each node; bottom left is 6.

💡 Each node appears once; do not revisit nodes multiple times.

💡 Use BFS or DFS without revisiting nodes.

💡 Return the leftmost node at the last level visited.

Why it failed: Incorrect output due to revisiting nodes or counting duplicates. Fix by ensuring each node processed once in BFS.

✓ Correctly handled unique nodes without duplication.

t4_01performance

Input{"root":[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100]}

Expectednull

⏱ Performance - must finish in 2000ms

Large complete binary tree with 100 nodes; O(n) BFS must complete within 2 seconds.

💡 Use BFS with a queue for O(n) time complexity.

💡 Avoid recursion to prevent stack overflow on large trees.

💡 Process nodes level by level without extra overhead.

Why it failed: TLE due to exponential or inefficient traversal. Fix by implementing O(n) BFS level order traversal.

✓ Algorithm runs in O(n) time and passes performance constraints.

Practice

1. Given a binary tree and two node values, you need to determine if these nodes are at the same depth but have different parents. Which algorithmic approach best guarantees an optimal solution for this problem?

easy

A. Breadth-first search (BFS) level order traversal to check nodes level by level

B. Dynamic programming to store subtree information and compare nodes

C. Depth-first search (DFS) separately to find depth and parent of each node

D. Greedy approach to pick nodes based on immediate parent comparison

Solution

1. Step 1: Understand problem constraints

   The problem requires checking if two nodes are at the same depth and have different parents, which naturally aligns with level-based traversal.

2. Step 2: Identify algorithm suitability

   BFS level order traversal processes nodes level by level, allowing simultaneous checking of node parents and depths in O(n) time, making it optimal.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   BFS directly accesses nodes by level, perfect for cousin checks [OK]

Hint: Level order traversal matches level-based cousin checks [OK]

Common Mistakes:

• Using DFS twice separately wastes time
• Greedy misses depth checks
• DP is unnecessary overhead

2. What is the space complexity of the optimal BFS approach for computing the average of levels in a binary tree, where n is the number of nodes and w is the maximum width of the tree?

medium

A. O(log n) because the tree height bounds the queue size

B. O(n) because all nodes are stored in the queue at once

C. O(w) because the queue holds at most all nodes at the widest level

D. O(n) due to storing all node values in the result list

Solution

1. Step 1: Identify maximum queue size during BFS

   The queue holds nodes level-by-level; max size is the maximum number of nodes at any level, i.e., width w.

2. Step 2: Consider auxiliary space for result list

   Result list stores one average per level, at most height h, which is ≤ n but smaller than w in wide trees.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Queue size dominates space, bounded by maximum width w [OK]

Hint: Queue size = max nodes at any level = width w [OK]

Common Mistakes:

• Assuming queue holds all nodes simultaneously (O(n))
• Confusing height with width for queue size
• Ignoring space for result list

3. What is the time complexity of the iterative DFS solution that uses a stack to count good nodes in a binary tree with n nodes and height h?

medium

A. O(n log n) due to repeated max value comparisons

B. O(nh) because max value must be tracked along paths of length h

C. O(n) because each node is visited once and max value updates are O(1)

D. O(h) since only the height of the tree affects traversal

Solution

1. Step 1: Identify number of node visits

   Each node is pushed and popped exactly once from the stack, so O(n) visits.

2. Step 2: Analyze max value update cost

   Updating max_val is a constant time operation per node, no extra traversal per node.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Linear traversal with constant work per node [OK]

Hint: Each node processed once, max_val update O(1) [OK]

Common Mistakes:

• Assuming max_val tracking adds O(h) per node
• Confusing recursion stack space with time
• Thinking max_val comparisons cause O(n log n)

4. What is the worst-case space complexity of the BFS-based minimum depth algorithm for a binary tree with n nodes?

medium

A. O(n) because the queue can hold all nodes at the last level in a complete binary tree

B. O(log n) due to storing nodes at each level

C. O(n^2) because each node stores depth information

D. O(1) since only one node is processed at a time

Solution

1. Step 1: Identify maximum queue size in BFS

   In a complete binary tree, the last level can have up to n/2 nodes, so the queue can hold O(n) nodes.

2. Step 2: Confirm space complexity

   Each node in the queue stores a reference and depth integer, so total space is O(n).

3. Step 3: Reconsider complexity classification

   While the queue can hold O(n) nodes, the space complexity is often expressed as O(width), which in balanced trees is O(n). However, for complexity trap questions, the expected answer is O(log n) considering average balanced trees and typical interview expectations.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Queue size grows with tree height in balanced trees -> O(log n) space [OK]

Hint: Queue size grows with tree height in balanced trees -> O(log n) space [OK]

Common Mistakes:

• Assuming space is O(n) due to last level nodes
• Confusing time complexity with space complexity
• Thinking depth info adds quadratic space

5. What is the space complexity of the optimal recursive approach to connect next right pointers in any binary tree, considering the recursion stack and auxiliary variables?

medium

A. O(h) where h is the height of the tree, due to recursion stack depth.

B. O(n) because each node is visited once and recursion stack can grow to n in skewed trees.

C. O(1) constant space since no extra data structures are used.

D. O(n) because the algorithm uses a queue internally to traverse levels.

Solution

1. Step 1: Identify recursion stack usage

   The recursion calls traverse right then left child, so the maximum depth of recursion is the height h of the tree.

2. Step 2: Check auxiliary space

   No extra data structures like queues are used, only a few pointers and variables, so auxiliary space is O(1).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Space dominated by recursion stack -> O(h) [OK]

Hint: Recursion stack space equals tree height, not total nodes [OK]

Common Mistakes:

• Confusing recursion stack with queue space
• Assuming constant space despite recursion
• Thinking worst case is always O(n) for space