Path Sum

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.

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool hasPathSum(TreeNode *root, int sum) {
13         
14         if(root == NULL)
15             return false;
16             
17         int tempSum = 0;
18         int flag = 0;
19         pathSum(root, tempSum, sum, flag);
20         if(flag == 1)
21             return true;
22         else
23             return false;
24     }
25     
26     void pathSum(TreeNode *root, int tempSum, int sum, int &flag)
27     {
28         if(root != NULL)
29         {
30             if(root->left == NULL && root->right == NULL)
31             {
32                 tempSum += root->val;
33                 if(tempSum == sum)
34                     flag = 1;
35                 
36                 return;
37             }
38         
39             else
40             {
41                 tempSum += root->val;
42                 pathSum(root->left, tempSum, sum, flag);
43                 pathSum(root->right, tempSum, sum, flag);
44             }
45         }
46     }
47 };

 

posted @ 2014-09-13 19:33  yqc  阅读(101)  评论(0)  编辑  收藏  举报