[leetcode] 124. Binary Tree Maximum Path Sum 解题报告

题目链接: https://leetcode.com/problems/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

.

思路: 使用DFS来搜索所有结点, 并且在搜索一个结点时返回从叶子结点到其左子树最大的路径和, 和从叶子结点到其右子树最大的路径和.

如果从左子树经过此结点到右子树的路径和大于最大值, 则更新最大值. 另外需要注意的是, 结点值竟然有负值, 所以在返回经过一个结点的最大路径和的时候有可能即不经过左子树也不经过右子树, 也就是路径不需要从叶子结点出发.

代码如下:

/**
* 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 DFS(TreeNode* root, int& ans)
{
if(root==NULL) return 0;
int left = DFS(root->left, ans);
int right = DFS(root->right, ans);
int val = left + right + root->val;
int Max = max(val, max(root->val, max(val-left, val-right)));
ans = max(ans, Max);
return max(root->val + max(left, right), root->val);
}
int maxPathSum(TreeNode* root) {
int ans = INT_MIN;
DFS(root, ans);
return ans;
}
};