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Binary Tree Maximum Path Sum

时间:2014-12-03 21:10:21      阅读:166      评论:0      收藏:0      [点我收藏+]

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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.

/**
 * 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 maxPathSum(TreeNode *root) {
        int tmp = help(root);
        return maxval;
    }
    int help(TreeNode *root){
        if(root == NULL) return 0;
        if(root->left == NULL && root->right == NULL){maxval = max(maxval, root->val); return root->val;}
        int lmax, rmax, ans, l, r, ret;
        lmax = rmax = ans = INT_MIN;
        l = r = 0;
        if(root->left) {
            l = help(root->left);
            lmax = max(l, l + root->val);
        }
        if(root->right) {
            r = help(root->right);
            rmax = max(r, r + root->val);
        }
        ans = max(lmax, rmax);
        ans = max(ans, root->val);
        ans = max(ans, l + r + root->val);
        maxval = max(ans, maxval);
        
        ret = max(l+root->val, r+root->val);
        ret = max(ret, root->val);
        return ret;
    }
    
    int maxval = INT_MIN;
};

 

Binary Tree Maximum Path Sum

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原文地址:http://www.cnblogs.com/code-swan/p/4141095.html

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