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Path Sum

时间:2014-12-03 21:12:00      阅读:153      评论:0      收藏:0      [点我收藏+]

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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 binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        bool flag = false;
        level(root,0,sum,flag);
        return flag;
    }
    void level(TreeNode *node, int tmpsum, int sum, bool& flag)  {
        if(flag) return;
        if(node == NULL) return;
        int newsum = tmpsum + node->val;
        if(newsum == sum && node->left == NULL && node->right == NULL){
            flag = true;
            return;
        }
        else{
            level(node->left, newsum, sum, flag);
            level(node->right, newsum, sum, flag);
        }
    }
};

 

Path Sum

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原文地址:http://www.cnblogs.com/code-swan/p/4141165.html

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