LeetCode OJ

题目：

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

For example:
Given the below binary tree,

       1
      / 
     2   3

Return 6.

解题思路：

采用后序遍历。设当前根节点的左子树的最大和为left_ans，右子树的最大和为right_ans;则ans 是 root->val + left_ans 、 root->val + right_ans 和 root->val + left_ans + right_ans的最大值。

而以该根节点的子树对父节点的贡献为 root->val + max(left_ans, right_ans);

代码：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int post_order(TreeNode *root, int &ans) {
        if (root == NULL) return 0;

        int left_ans = 0, right_ans = 0;
        if (root->left) left_ans = max(left_ans, post_order(root->left, ans));
        
        if (root->right) right_ans = max(right_ans, post_order(root->right, ans));
        
        ans = max(ans, root->val + left_ans + right_ans);
        return max(root->val + left_ans, root->val + right_ans);
    }
    int maxPathSum(TreeNode *root) {
        if (root == NULL) return 0;
        
        int ans = root->val;
        post_order(root, ans);
        
        return ans;
    }
};