https://leetcode-cn.com/problems/binary-tree-maximum-path-sum/
思路:
d
f
s
dfs
dfs就完事了,每个节点维护一个从叶子节点到当前节点的路径之和的最大值,有点
d
p
dp
dp的感觉。
/**
* 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 ans;
int maxPathSum(TreeNode* root) {
ans=root->val;
dfs(root);
return ans;
}
int dfs(TreeNode* root){
int l=0,r=0;
if(root->left!=NULL)
l=dfs(root->left);
if(root->right!=NULL)
r=dfs(root->right);
ans=max(ans,l+r+root->val);
int tmp=max(l,r)+root->val;
return max(tmp,0);
}
};