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

 /**
  * 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 maxVal = INT_MIN;
     int maxPathSum(TreeNode* root) {
         if (root == NULL)
             ;

         maxSum(root);
         return maxVal;
     }

     /*递归函数*/
     int maxSum(TreeNode *root)
     {
         if (root == NULL)
             ;

         /*求以root为根的当前子树的最大路径和*/
         int curVal = root->val;
         int lmaxSum = maxSum(root->left), rmaxSum = maxSum(root->right);
         )
             curVal += lmaxSum;
         )
             curVal += rmaxSum;

         if (curVal > maxVal)
             maxVal = curVal;

         /*返回以当前root为根的子树的最大路径和*/
         return max(root->val, max(root->val + lmaxSum, root->val + rmaxSum));
     }
 };

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