leetcode-124-Binary Tree Maximum Path Sum

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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.

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Example 1:

Input: [1,2,3]

       1
      / \
     2   3

Output: 6
Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

对于每条路径，都有一个最高节点，每个节点都可以看作一个最高节点，所有最高节点中的最大值即为所求。因此可以递归求解，对于每个最高节点，路径肯定包含其节点的值，左或右子树的最大路径不能为负，否则取0，分别左右子树的最大路径和最高节点的值累加，即为以最高节点为根的最大路径和，更新maxSum;另外对于每个最高节点，只能返回路径值较大一侧的子树给上一层，否则不符题意。

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/**
 * 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 maxPathSum(TreeNode* root) {
        maxPathDown(root);
        return ans;
    }
    int maxPathDown(TreeNode* root){
        if(root==NULL) return 0;
        int left=max(0,maxPathDown(root->left));
        int right=max(0,maxPathDown(root->right));
        ans=max(ans,root->val+left+right);
        return root->val+max(0,max(left,right));
    }
private:
    int ans=INT_MIN;
    
};
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