【LeetCode】124. Binary Tree Maximum Path Sum

Binary Tree Maximum Path Sum

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.

树结构显然用递归来解，解题关键：

1、对于每一层递归，只有包含此层树根节点的值才可以返回到上层。否则路径将不连续。

2、返回的值最多为根节点加上左右子树中的一个返回值，而不能加上两个返回值。否则路径将分叉。

在这两个前提下有个需要注意的问题，最上层返回的值并不一定是满足要求的最大值，

因为最大值对应的路径不一定包含root的值，可能存在于某个子树上。

因此解决方案为设置全局变量maxSum，在递归过程中不断更新最大值。

/*** Definition for binary tree* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:int maxSum;Solution(){maxSum = INT_MIN;}int maxPathSum(TreeNode* root){Helper(root);return maxSum;}int Helper(TreeNode *root) {if(!root)return INT_MIN;else{int left = Helper(root->left);int right = Helper(root->right);if(root->val >= 0){//allways include rootif(left >= 0 && right >= 0)maxSum = max(maxSum, root->val+left+right);else if(left >= 0 && right < 0)maxSum = max(maxSum, root->val+left);else if(left < 0 && right >= 0)maxSum = max(maxSum, root->val+right);elsemaxSum = max(maxSum, root->val);}else{if(left >= 0 && right >= 0)maxSum = max(maxSum, max(root->val+left+right, max(left, right)));else if(left >= 0 && right < 0)maxSum = max(maxSum, left);else if(left < 0 && right >= 0)maxSum = max(maxSum, right);elsemaxSum = max(maxSum, max(root->val, max(left, right)));}//return only one path, do not add left and right at the same timereturn max(root->val+max(0, left), root->val+max(0, right));}}
};

转载于:https://www.cnblogs.com/ganganloveu/p/4126953.html

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