124. Binary Tree Maximum Path Sum *HARD* — 二叉树中节点和最大的路径的节点和

Given a 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 does not need to go through the root.

For example:
Given the below binary tree,

       1
      / \
     2   3

Return 6.

/**
 * 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 getMax(TreeNode* root, int &ans)
    {
        if(!root)
            return ;
        int leftMax = getMax(root->left, ans), rightMax = getMax(root->right, ans), Max = max(leftMax, rightMax);
        int curMax = ((leftMax > ) ? leftMax : ) + ((rightMax > ) ? rightMax : ) + root->val;
        if(curMax > ans)
            ans = curMax;
        if(Max <= )
            return root->val;
        return Max + root->val;
    }
    int maxPathSum(TreeNode* root) {
        int ans = INT_MIN;
        getMax(root, ans);
        return ans;
    }
};

分别求左右子树的最大节点和，若为负，则忽略，若为正，则加上本身的节点值。