二叉树--二叉树最长路径

124. Binary Tree Maximum Path Sum

Given a non-empty binary tree, find the maximum path sum.

For this problem, 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.

Example 1:

Input: [1,2,3] 1 / \ 2 3Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]   -10    / \   9  20     /  \    15   7Output: 42

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int MaxPathLength = -0x3f3f3f3f;
    int maxPathSum(TreeNode* root){
        if(root == nullptr){
            return 0;
        }
        maxSum(root);
        return MaxPathLength;
    }
    int maxSum(TreeNode* root){
        if(root == nullptr){
            return 0;
        }
        int CurNode = root->val;
        int LeftNode = maxSum(root->left);
        int RightNode = maxSum(root->right);
        if(LeftNode > 0){
            CurNode += LeftNode;
        }
        if(RightNode > 0){
            CurNode += RightNode;
        }
        if(CurNode > MaxPathLength){
            MaxPathLength = CurNode;
        }
        return max(root->val,max(root->val + LeftNode,root->val + RightNode));
    }
};