Binary Tree Maximum Path Sum -- leetcode

标签：二叉树   path   leetcode   深度优先   

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.

基本思路：

路径为，左子树的深度+根结点值+右子树的深度。

对每一个结点，都作如上运算，并把路径最大值存在一个全局变量中。

另外，此题，由于结点值存在负数。当树深度为负时，需要舍弃。因为对最长路径没有帮助。

/**
 * 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 maxPathSum(TreeNode* root) {
        int max_path = INT_MIN;
        dfs(root, max_path);
        return max_path;
    }
    
    int dfs(TreeNode *root, int &max_path) {
        if (!root) return 0;
        int left = dfs(root->left, max_path);
        int right = dfs(root->right, max_path);
        max_path = max(max_path, left+right+root->val);
        return max(0, max(left, right)+root->val);
    }
};

Binary Tree Maximum Path Sum -- leetcode

标签：二叉树   path   leetcode   深度优先   

原文地址：http://blog.csdn.net/elton_xiao/article/details/45790657