Binary Tree Maximum Path Sum

问题
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

解答

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution1 {
public:
int maxPathSum(TreeNode* root)
{
if(!root) return 0;
if(!root->left && !root->right) return root->val;
else
{
int s1=maxPathSum(root->left);
int s2=maxPathSum(root->right);

max=0;
path(root->left,0);
int s3=max;

max=0;
path(root->right,0);
int s4=max;

if(s3+s4+root->val>s1 && s3+s4+root->val>s2)
return s3+s4+root->val;
else
{
return (s1>s2)?s1:s2;
}
}
}

public:
void path(TreeNode *root,int sum)
{
if(root)
{
if(root->left==NULL && root->right==NULL)
{
//a.push_back(root->val);
sum+=root->val;
if(sum>max)
max=sum;
}
else
{
sum+=root->val;
path(root->left,sum);
path(root->right,sum);
}
}
}
public:
int max;
};

/**
* 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 ans;
int scanT(TreeNode* root){
if(root == NULL) return 0;
int left = scanT(root -> left);
int right = scanT(root -> right);
int val = root -> val;
if(left > 0) val += left;
if(right > 0) val += right;
if(val > ans) ans = val;
return max(root->val ,max(left +  root -> val , right + root -> val));
}
int maxPathSum(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(root == NULL) return 0;
ans = root -> val;
scanT(root);
return ans;
}
};