LeetCode:Binary Tree Maximum Path Sum
题目如下： 二叉树的最大路径和（题目链接）
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.
分析：设f(p) 是以节点p作为路径端点的最大路径和，例如下面的树
f(2) = 2+5=7, f(3) = 3+6=9, f(1) = f(3)+1=10
即f(p) = max(p->val , p->val+f(p->left), p->val+f(p->right)), 特别的当p=NULL 时，f(p) = 0; 本文地址
有以上可知经过p节点的路径的最大和maxsum(p) = max(f(p), f(p->left)+f(p->right)+p->val), 即取p为端点和p为中间节点的路径的较大值
因此可以通过树的后序遍历求得每个节点的maxsum，选取最大的即为最终结果，代码如下：
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int maxPathSum(TreeNode *root) { 13 // IMPORTANT: Please reset any member data you declared, as 14 // the same Solution instance will be reused for each test case. 15 if(root == NULL)return 0; 16 int res = INT_MIN; 17 postorderTrav(root, res); 18 return res; 19 } 20 int postorderTrav(TreeNode *root, int& res) 21 {//返回值：root作为端点的path的最大值 22 int left = 0, right = 0; 23 if(root->left)left = postorderTrav(root->left, res); 24 if(root->right)right = postorderTrav(root->right, res); 25 int res_this = max(max(left+root->val, right+root->val), root->val); 26 int tmp = max(res_this, left + right + root->val); 27 if(tmp > res) 28 res = tmp; 29 return res_this; 30 } 31 };
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