124. Binary Tree Maximum Path Sum

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124. Binary Tree Maximum Path Sum

Description

A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once . Note that the path does not need to pass through the root.

The path sum of a path is the sum of the node’s values in the path.

Given the root of a binary tree, return the maximum path sum of any non-empty path.

Example 1:

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Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.

Example 2:

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Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.

Constraints:

  • The number of nodes in the tree is in the range [1, 3 * 10^4].
  • -1000 <= Node.val <= 1000

Hints/Notes

Solution

Language: C++

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    int res = INT_MIN;

    int maxPathSum(TreeNode* root) {
        traverse(root);

        return res;
    }

    int traverse(TreeNode* root) {
        int ans = 0;
        if (!root) {
            return ans;
        }

        ans += root->val;

        int left = traverse(root->left);
        int right = traverse(root->right);

        res = max(res, root->val + left + right);

        ans += max(left, right);

        return max(0, ans);
    }
};