124. 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 must contain at least one node and does not need to go through the root.
解題思路：
樹結構顯然用遞迴來解，解題關鍵：
1、對於每一層遞迴，只有包含此層樹根節點的值才可以返回到上層。否則路徑將不連續。
2、返回的值最多為根節點加上左右子樹中的一個返回值，而不能加上兩個返回值。否則路徑將分叉。
在這兩個前提下有個需要注意的問題，最上層返回的值並不一定是滿足要求的最大值，
因為最大值對應的路徑不一定包含root的值，可能存在於某個子樹上。
因此解決方案為設定全域性變數maxSum，在遞迴過程中不斷更新最大值。
實驗程式碼：
/**
* 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 maxSum;
Solution()
{
maxSum = INT_MIN;
}
int maxPathSum(TreeNode* root)
{
Helper(root);
return maxSum;
}
int Helper(TreeNode *root) {
if(!root)
return INT_MIN;
else
{
int left = Helper(root->left);
int right = Helper(root->right);
if(root->val >= 0)
{//allways include root
if(left >= 0 && right >= 0)
maxSum = max(maxSum, root->val+left+right);
else if(left >= 0 && right < 0)
maxSum = max(maxSum, root->val+left);
else if(left < 0 && right >= 0)
maxSum = max(maxSum, root->val+right);
else
maxSum = max(maxSum, root->val);
}
else
{
if(left >= 0 && right >= 0)
maxSum = max(maxSum, max(root->val+left+right, max(left, right)));
else if(left >= 0 && right < 0)
maxSum = max(maxSum, left);
else if(left < 0 && right >= 0)
maxSum = max(maxSum, right);
else
maxSum = max(maxSum, max(root->val, max(left, right)));
}
//return only one path, do not add left and right at the same time
return max(root->val+max(0, left), root->val+max(0, right));
}
}
};