题目
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.
分析
给定一颗二叉树，求其最大路径和。对于二叉树，算法大多可以选择递归解决，此题也不例外。
如果只是一个节点,那么当然就是这个节点的值了.
如果这个作为root,那么最长路应该就是..
F(left) + F(right) + val...当然如果left,或者right<0就不用加了的= =
从下往上递归遍历...
如果不这个不是root,那么就不能把left和right加起来了...因为只是一条路...
AC代码
1 /** 2 * Definition for a binary tree node. 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 11 class Solution { 12 public: 13 int maxVal = INT_MIN; 14 int maxPathSum(TreeNode* root) { 15 if (root == NULL) 16 return 0; 17 18 maxSum(root); 19 return maxVal; 20 } 21 22 /*递归函数*/ 23 int maxSum(TreeNode *root) 24 { 25 if (root == NULL) 26 return 0; 27 28 /*求以root为根的当前子树的最大路径和*/ 29 int curVal = root->val; 30 int lmaxSum = maxSum(root->left), rmaxSum = maxSum(root->right); 31 if (lmaxSum > 0) 32 curVal += lmaxSum; 33 if (rmaxSum > 0) 34 curVal += rmaxSum; 35 36 if (curVal > maxVal) 37 maxVal = curVal; 38 39 /*返回以当前root为根的子树的最大路径和*/ 40 return max(root->val, max(root->val + lmaxSum, root->val + rmaxSum)); 41 } 42 };