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LeetCode 112:Path Sum

时间:2015-12-22 13:17:54      阅读:136      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.

 //利用深度优先搜索DFS,当搜索到的节点为叶子节点时,才判断总和是否等于给定的sum值
class Solution {
public:
	bool dfs(TreeNode* node, int sum, int curSum){
		if (node == NULL)
			return false;
		if (node->left == NULL && node->right == NULL)
			return curSum + node->val == sum;

		return dfs(node->left, sum, curSum + node->val) || dfs(node->right, sum, curSum + node->val);
	}

	bool hasPathSum(TreeNode* root, int sum) {
		return dfs(root, sum, 0);
	}
};

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LeetCode 112:Path Sum

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原文地址:http://blog.csdn.net/geekmanong/article/details/50379167

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