【树形dp】Binary Tree Maximum Path Sum

Binary Tree Maximum Path Sum

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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

class Solution {
public:
int maxPathSum(TreeNode *root) {
int maxV=0x80000000;////@@@@error: maxV=0; then the negative path sum will be blocked ,since they are smaller than 0.
dfs(root,maxV);
return maxV;
}
int dfs(TreeNode *node,int &gMaxV){
int lv=0,rv=0;
if(node->left!=NULL) lv=dfs(node->left,gMaxV);//@@@error:no matching function for call to 'Solution::dfs(left)'
if(node->right!=NULL) rv=dfs(node->right,gMaxV);
gMaxV=max(gMaxV,lv+rv+node->val);
return max(0,max(lv,rv)+node->val);
}
};