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 andsum = 22
,
return true, as there exist a root-to-leaf path
5->4->11->2
which sum is 22.
这个题目非常适合用递归来做,C++代码实现非常简单:
bool hasPathSum(TreeNode *root, int sum) {
if (!root) return false;
if (root->left && hasPathSum(root->left, sum - root->val)) return true;
if (root->right && hasPathSum(root->right, sum - root->val)) return true;
if (root->left == nullptr && root->right == nullptr && root->val == sum) return true;
return false;
}
时间性能如下图所示:
原文:http://blog.csdn.net/chfe007/article/details/43453597