[LeetCode] PathSum

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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.

思路：DFS.时间复杂度O(n), 空间复杂度O(lgN)

相关题目：《剑指offer》面试题25

 1 class Solution {
 2 public:
 3     bool hasPathSum(TreeNode *root, int sum) {
 4         if (root == NULL) {
 5             return false;
 6         }
 7         
 8         if (root->left == NULL && root->right == NULL) {
 9             return root->val == sum;
10         }
11         
12         return hasPathSum(root->left, sum - (root->val)) ||
13                 hasPathSum(root->right, sum - (root->val));
14     }
15 };

[LeetCode] PathSum

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原文地址：http://www.cnblogs.com/vincently/p/4130398.html