124. Binary Tree Maximum Path Sum

124. Binary Tree Maximum Path Sum

problem description

Given a non-empty 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.

Example 1:

Input: [1,2,3]

       1
      / \
     2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

algorithm thought

还是典型的树问题，使用递归解决。

这里题目有点描述不清楚，这里只有根节点可以连接两个子树，也就是加上子树上计算的结果。但是如果是中间节点，那就只能连接一个子树，选择左子树或者右子树。这里当然是选择最大的一个子树。

code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int res;
    int maxtosum(TreeNode* root){
        if(root==NULL)
            return 0;
         int l=maxtosum(root->left);
         int r=maxtosum(root->right);
        if(l<0) l=0;
        if(r<0) r=0;
        if(root->val+l+r>res) res=root->val+l+r;
        return root->val+=max(l,r);
    }
    int maxPathSum(TreeNode* root) {
        res=INT_MIN;
        maxtosum(root);
        return res;
    }
};

algorithm analysis

对所有的节点遍历一遍得出结果，时间复杂度O(n)