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Path Sum

时间:2015-10-17 17:39:25      阅读:193      评论:0      收藏:0      [点我收藏+]

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             /             4   8
           /   /           11  13  4
         /  \              7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

/**
 * 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:
    bool execute(TreeNode* root,int sum){
        sum-=root->val;
        if(root->left==NULL&&root->right==NULL){
            if(sum==0)
                return true;
        }
        if(root->left&&execute(root->left,sum)){
            return true;
        }
        if(root->right&&execute(root->right,sum)){
            return true;
        }
        return false;
    }
    
    bool hasPathSum(TreeNode* root, int sum) {
        if(root==NULL)
            return false;
        return execute(root,sum); 
    }
};

 

Path Sum

原文:http://www.cnblogs.com/michaelzhao10/p/4887640.html

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