【LeetCode】124. Binary Tree Maximum Path Sum

Binary Tree Maximum Path Sum

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

Return 6.

树结构显然用递归来解，解题关键：

1、对于每一层递归，只有包含此层树根节点的值才可以返回到上层。否则路径将不连续。

2、返回的值最多为根节点加上左右子树中的一个返回值，而不能加上两个返回值。否则路径将分叉。

在这两个前提下有个需要注意的问题，最上层返回的值并不一定是满足要求的最大值，

因为最大值对应的路径不一定包含root的值，可能存在于某个子树上。

因此解决方案为设置全局变量maxSum，在递归过程中不断更新最大值。

/**
 * 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));
        }
    }
};