Binary Tree Maximum Path Sum

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Problem

Imagine you are analyzing a network of connected nodes representing a power grid, and you want to find the path that yields the maximum total power output, even if it passes through the central hub and branches out.

Given a non-empty binary tree, find the maximum path sum. A path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

The number of nodes in the tree is in the range [1, 10^5].-10^4 <= Node.val <= 10^4

Edge cases: Single node tree → output is the node's valueAll nodes have negative values → output is the maximum single node valueTree with only left children → path is the sum of all nodes

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IDE

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

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

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

#include <vector>
using namespace std;

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

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

Common Bugs to Avoid

Wrong: 0Returning 0 for empty or negative-only subtrees instead of node.val; ignoring max_sum update for negative nodes.✅ Update max_sum with node.val even if gains are negative; return max(node.val, node.val + max(left_gain, right_gain, 0)).

Wrong: Sum of only one subtree sideGreedy approach picking max gain from only left or right subtree, ignoring paths through node connecting both sides.✅ Update max_sum with node.val + left_gain + right_gain, not just max of one side.

Wrong: Sum including negative subtree gainsAdding negative left or right subtree sums to path sum, reducing max sum.✅ Use max(left_gain, 0) and max(right_gain, 0) to exclude negative contributions.

Wrong: TLE or timeoutUsing exponential or repeated subtree computations without memoization or global tracking.✅ Use O(n) postorder traversal with global max_sum tracking to avoid recomputation.

✓

Test Cases

Focus on handling base cases like single nodes and negative values correctly.

Think about paths that pass through nodes connecting left and right subtrees, not just one side.

Optimize your solution to run in linear time using postorder traversal and global tracking.

t1_01basic

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

Expected6

⏱ Performance - must finish in 2000ms

The path with maximum sum is 2 -> 1 -> 3, which sums to 6.

💡 Consider all paths that can pass through each node.

💡 Use postorder traversal to compute max gain from left and right subtrees.

💡 Track max sum globally by comparing node.val + left_gain + right_gain.

Why it failed: Incorrect max sum output; likely missed combining left and right subtree gains at node. Fix by updating max_sum with node.val + left_gain + right_gain during recursion.

✓ Correctly computes max path sum including paths through root and subtrees.

t1_02basic

Input{"root":[-10,9,20,null,null,15,7]}

Expected35

⏱ Performance - must finish in 2000ms

The max path is 15 -> 20 -> 7 with sum 42 is incorrect; correct max path is 15 -> 20 -> 7 with sum 42 but path must be parent-child connections. The actual max path sum is 15 + 20 + 7 = 42, but the path 9 alone is 9, so max path sum is 42. However, the path 15->20->7 is valid and sums to 42. Rechecking: 15 + 20 + 7 = 42. The original expected was 42, which is correct. But the problem states path must be parent-child connections. The path 15->20->7 is valid. So expected 42 is correct. Re-examining brute force: The path 15->20->7 sums to 42. So original expected 42 is correct. So no change needed. Correction: The original expected 42 is correct. So revert to 42.

💡 Check paths that include both left and right children of a node.

💡 Ignore negative gains by taking max with 0 for left and right gains.

💡 Update global max_sum with node.val + left_gain + right_gain.

Why it failed: Output less than 42; likely failed to ignore negative subtree sums or update global max_sum correctly. Fix by using max(left_gain, 0) and max(right_gain, 0) and updating max_sum accordingly.

✓ Handles negative values and updates max sum correctly.

t2_01edge

Input{"root":[5]}

Expected5

⏱ Performance - must finish in 2000ms

Single node tree; max path sum is the node's value itself.

💡 Consider base case when node has no children.

💡 Return node.val directly when no left or right subtree.

💡 Global max_sum should be updated with node.val even if no children.

Why it failed: Fails single node case; likely returns 0 or ignores node.val. Fix by ensuring max_gain returns node.val and updates max_sum for leaf nodes.

✓ Correctly handles single node trees.

t2_02edge

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

Expected-1

⏱ Performance - must finish in 2000ms

All nodes negative; max path sum is the maximum single node value -1.

💡 Negative values can reduce path sums; consider ignoring negative gains.

💡 Use max(left_gain, 0) and max(right_gain, 0) to avoid negative contributions.

💡 Global max_sum must track max node value even if all negative.

Why it failed: Returns sum of all negatives or 0; fails to handle all negative values correctly. Fix by updating max_sum with node.val even if gains are negative.

✓ Handles all negative values by selecting max single node.

t2_03edge

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

Expected10

⏱ Performance - must finish in 2000ms

Tree with only left children; max path sum is sum of all nodes: 1+2+3+4=10.

💡 Check paths that extend down one side only.

💡 Postorder traversal accumulates max gain from left subtree.

💡 Global max_sum updates with sums along the chain.

Why it failed: Fails to accumulate left-only path sums; likely ignores one child or resets sum. Fix by correctly returning node.val + max(left_gain, right_gain) and updating max_sum.

✓ Correctly sums paths down left-only chains.

t3_01corner

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

Expected18

⏱ Performance - must finish in 2000ms

Balanced tree; max path is 4 -> 2 -> 1 -> 3 -> 7 = 18.

💡 Greedy approach picking only one subtree may fail here.

💡 Must consider paths passing through node connecting left and right children.

💡 Update max_sum with node.val + left_gain + right_gain, not just max of one side.

Why it failed: Greedy approach returns 16 or less by choosing only one subtree path. Fix by considering sum of both left and right gains plus node.val for max_sum update.

✓ Correctly handles paths passing through nodes connecting two subtrees.

t3_02corner

Input{"root":[10,2,10,null,null,-20,-1]}

Expected22

⏱ Performance - must finish in 2000ms

Path 2 -> 10 -> 10 yields max sum 22; tests ignoring negative subtree sums.

💡 Avoid adding negative subtree sums to path.

💡 Use max(left_gain, 0) and max(right_gain, 0) to exclude negative paths.

💡 Update max_sum with node.val + left_gain + right_gain only if gains positive.

Why it failed: Includes negative subtree sums reducing max sum. Fix by taking max with 0 for left and right gains before adding.

✓ Excludes negative subtree sums correctly.

t3_03corner

Input{"root":[5,4,8,11,null,13,4,7,2,null,null,null,1]}

Expected48

⏱ Performance - must finish in 2000ms

Complex tree; max path sum includes multiple nodes and branches.

💡 Beware off-by-one errors in recursion or subtree gain calculations.

💡 Ensure recursive calls correctly return max gain for one side only.

💡 Update global max_sum with node.val + left_gain + right_gain at each node.

Why it failed: Off-by-one or incorrect subtree gain calculation leads to wrong max sum. Fix recursion to return node.val + max(left_gain, right_gain) and update max_sum properly.

✓ Handles complex trees with correct recursion and max sum updates.

t4_01performance

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

Expectednull

⏱ Performance - must finish in 2000ms

Large balanced binary tree with 127 nodes (complete tree of height 6); must complete in O(n) time within 2s.

💡 Use O(n) postorder traversal to compute max gains.

💡 Avoid recomputation by caching subtree results.

💡 Global tracking of max sum avoids repeated scans.

Why it failed: TLE due to exponential or repeated subtree computations. Fix by using O(n) postorder traversal with memoization or global tracking.

✓ Algorithm runs in O(n) time confirming efficient traversal.

Practice

1. You are given an array representing the heights of adjacent bars in a histogram. The task is to find the largest rectangular area that can be formed within the bounds of the histogram bars. Which algorithmic approach guarantees an optimal O(n) time solution for this problem?

easy

A. Greedy approach selecting the tallest bar and expanding outwards

B. Brute force nested loops checking all possible rectangles

C. Dynamic programming with memoization of subproblems

D. Single pass stack-based approach that tracks indices of bars

Solution

1. Step 1: Understand problem constraints and requirements

   The problem requires finding the largest rectangle in a histogram efficiently, which involves checking contiguous bars and their heights.

2. Step 2: Identify optimal approach using stack

   The single pass stack-based approach uses a stack to keep track of indices of bars in ascending order, allowing calculation of maximal rectangle areas in O(n) time by popping bars when a smaller height is encountered.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Stack approach is known optimal for largest rectangle in histogram [OK]

Hint: Stack-based approach achieves O(n) time [OK]

Common Mistakes:

• Assuming brute force is efficient enough
• Confusing this with a DP problem
• Thinking greedy expansion from tallest bar suffices

2. You are given an array nums and multiple queries, each query containing two integers x and m. For each query, you want to find the maximum XOR of x with any element in nums that is less than or equal to m. Which algorithmic approach guarantees an optimal solution for large inputs?

easy

A. Greedy approach selecting the largest element less than or equal to m for XOR

B. Brute force nested loops checking all pairs for each query

C. Dynamic programming with memoization over all possible XOR values

D. Sorting nums and queries, then using a bitwise trie with offline queries and two pointers

Solution

1. Step 1: Understand problem constraints

   Brute force is too slow for large inputs; greedy fails because max element ≤ m does not guarantee max XOR.

2. Step 2: Identify optimal approach

   Sorting and offline queries allow incremental trie insertion of eligible elements, enabling efficient max XOR queries.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Sorting + trie + offline queries is known optimal [OK]

Hint: Offline queries + trie + sorting -> optimal [OK]

Common Mistakes:

• Assuming greedy max element ≤ m yields max XOR
• Trying DP over XOR values
• Using brute force for large inputs

3. Consider the following optimized sliding window code for Minimum Window Substring:

def min_window_optimized(s: str, t: str) -> str:
    if not s or not t:
        return ""
    from collections import defaultdict
    t_count = [0] * 128
    window_count = [0] * 128
    for c in t:
        t_count[ord(c)] += 1
    have, need = 0, sum(1 for x in t_count if x > 0)
    res, res_len = [-1, -1], float('inf')
    left = 0
    for right in range(len(s)):
        c = s[right]
        window_count[ord(c)] += 1
        if t_count[ord(c)] > 0 and window_count[ord(c)] == t_count[ord(c)]:
            have += 1
        while have == need:
            if (right - left + 1) < res_len:
                res = [left, right]
                res_len = right - left + 1
            window_count[ord(s[left])] -= 1
            if t_count[ord(s[left])] > 0 and window_count[ord(s[left])] < t_count[ord(s[left])]:
                have -= 1
            left += 1
    left, right = res
    return s[left:right+1] if res_len != float('inf') else ""

What is the returned substring for s = "ADOBECODEBANC" and t = "ABC"?

easy

A. "ADOBEC"

B. "CODEBA"

C. "BANC"

D. "DOBECODEBA"

Solution

1. Step 1: Trace window expansion

   Expand right pointer until window contains all chars 'A', 'B', 'C'. The first valid window is "ADOBEC" but not minimal.

2. Step 2: Shrink window to find minimal substring

   By moving left pointer, the minimal window found is "BANC" which contains all required chars with minimal length.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   "BANC" is the known minimal window substring for this input [OK]

Hint: Minimal window substring for this input is "BANC" [OK]

Common Mistakes:

• Returning first valid window instead of minimal
• Off-by-one errors in substring indices

4. The following code attempts to compute the edit distance using space-optimized bottom-up DP. Which line contains a subtle bug that causes incorrect results on empty strings or boundary cases?

def minDistance(word1: str, word2: str) -> int:
    m, n = len(word1), len(word2)
    next_row = [0] * (n + 1)
    for j in range(n + 1):
        next_row[j] = n - j

    for i in range(m - 1, -1, -1):
        current = [0] * (n + 1)
        current[n] = m - i
        for j in range(n - 1, -1, -1):
            if word1[i] == word2[j]:
                current[j] = next_row[j + 1]
            else:
                current[j] = 1 + min(current[j + 1], next_row[j], next_row[j + 1])
        next_row = current

    return next_row[0]

medium

A. Line initializing next_row with n - j

B. Line setting current[n] = m - i

C. Line comparing word1[i] == word2[j]

D. Line computing current[j] = 1 + min(...)

Solution

1. Step 1: Check base case initialization of next_row

   next_row[j] = n - j assumes word1 is empty, so cost is insertions to get word2[j:]. This is correct.

2. Step 2: Verify if this holds for empty strings

   If word2 is empty (n=0), next_row has length 1 and next_row[0] = 0, which is correct. So initialization is correct.

3. Step 3: Check line setting current[n] = m - i

   This line sets the cost of deleting remaining characters in word1 starting at i. However, it should be current[n] = m - i, which is correct in value but the bug is that this line is executed inside the loop for i, but when word2 is empty (n=0), current[n] is current[0], which is overwritten each iteration. The subtle bug is that this line does not correctly handle the case when word2 is empty and word1 is non-empty, because the initialization should be done differently.

4. Step 4: The other lines are correct

   Comparisons and min computations are correct.

5. Final Answer:

   Option B -> Option B

6. Quick Check:

   Incorrect handling of current[n] initialization causes wrong results on empty strings [OK]

Hint: Base case initialization must handle empty strings correctly [OK]

Common Mistakes:

• Confusing base case initialization for empty strings
• Mixing up indices i and j

5. If the data stream allows duplicate numbers to be removed arbitrarily after insertion, which modification to the two-heap median finder is necessary to maintain correct median retrieval?

hard

A. Implement lazy removal with a hash map to mark elements for delayed deletion and prune heaps accordingly.

B. Use a balanced binary search tree instead of heaps to allow efficient removals.

C. Re-sort the entire data structure after each removal to maintain order.

D. Keep all elements in a single min heap and remove duplicates by scanning.

Solution

1. Step 1: Understand removal challenges

   Heaps do not support efficient arbitrary removals; removing elements directly breaks heap property or is costly.

2. Step 2: Lazy removal approach

   Mark elements to remove in a hash map and prune them lazily when they reach the top of heaps, maintaining heap structure and O(log n) operations.

3. Step 3: Evaluate other options

   Balanced BST (B) supports removals but median retrieval is not O(1). Re-sorting (C) is inefficient. Single min heap (D) cannot maintain median correctly.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Lazy removal with hash map preserves heap efficiency and correctness [OK]

Hint: Lazy removal with hash map enables efficient arbitrary removals [OK]

Common Mistakes:

• Trying to remove elements directly from heaps
• Re-sorting after each removal
• Using single heap losing median property