LeetCode No.124 Binary Tree Maximum Path Sum

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.

========================================================================

题目链接：https://leetcode.com/problems/binary-tree-maximum-path-sum/

题目大意：求出二叉树中任一路径和的最大值，路径的起点和终点可以为任意节点。

思路：利用递归的做法，先初始化ans = -2147483648，maxPath函数能算出root为根节点的路径最大值以及从root出发（非根节点）的路径最大值。

参考代码：

/** * 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) {        if ( root == NULL )            return 0 ;        int ans = -2147483648 ;//初始化为int值最小        maxPath( root , ans ) ;//计算        return ans ;    }    int maxPath ( TreeNode* root , int& ans )    {        if ( root == NULL )            return 0 ;        int l = maxPath ( root -> left , ans ) , r = maxPath ( root -> right , ans ) ;//计算左右最大值        int sum = root -> val ;        if ( l > 0 )            sum += l ;        if ( r > 0 )            sum += r ;        ans = max ( ans , sum ) ;//root节点为头结点，更新结果        return max ( root -> val , max ( root -> val + l , root -> val + r ) ) ;//返回最大值    }};