# LeetCode 124:Binary Tree Maximum Path Sum

Datetime:2016-08-23 01:23:14         Topic: LeetCode          Share        Original >>

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` .

Subscribe to see which companies asked this question

```//找出二叉树任意一点到另一点的路径,使得和最大.
//解题思路：后序遍历，先计算左右子树的值l和r，若l<0或r<0，则不用加上
class Solution {
public:
int maxSum;
int maxPathSum(TreeNode* root) {
if (root == NULL) return 0;
maxSum =root->val;
dfs(root);
return maxSum;
}

int dfs(TreeNode* root){
if (root == NULL)
return 0;
int l = dfs(root->left);
int r = dfs(root->right);
int sum = root->val;
if (l > 0) sum = sum + l;
if (r > 0) sum = sum + r;
maxSum = max(maxSum, sum);
return max(root->val, max(l + root->val, r + root->val));
}
};```
```#include<iostream>
#include<new>
#include<vector>
#include<algorithm>
using namespace std;

//Definition for binary tree
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
//找出二叉树任意一点到另一点的路径,使得和最大.
//解题思路：后序遍历，先计算左右子树的值l和r，若l<0或r<0，则不用加上
class Solution {
public:
int maxSum;
int maxPathSum(TreeNode* root) {
if (root == NULL) return 0;
maxSum =root->val;
dfs(root);
return maxSum;
}

int dfs(TreeNode* root){
if (root == NULL)
return 0;
int l = dfs(root->left);
int r = dfs(root->right);
int sum = root->val;
if (l > 0) sum = sum + l;
if (r > 0) sum = sum + r;
maxSum = max(maxSum, sum);
return max(root->val, max(l + root->val, r + root->val));
}
};
void createTree(TreeNode *&root)
{
int i;
cin >> i;
if (i != 0)
{
root = new TreeNode(i);
if (root == NULL)
return;
createTree(root->left);
createTree(root->right);
}
}

int main()
{
Solution s;
TreeNode *root;
createTree(root);
int sum = s.maxPathSum(root);
cout << sum << endl;
system("pause");
return 0;
}```