Path Sum **

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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.

Analyse: DFS.

1. Recursion

    Runtime: 12ms.

 1 /**
 2  * Definition for a binary tree node.
 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         if(!root) return false;
14         if(!root->left && !root->right) return sum == root->val;
15         
16         return hasPathSum(root->left, sum - root->val) ||
17                hasPathSum(root->right, sum - root->val);
18     }
19 };

2. Iteration-------need to think about a solution....

    Runtime: 

Path Sum **

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原文地址：http://www.cnblogs.com/amazingzoe/p/4692715.html