LeetCode----Path Sum

Path Sum Total Accepted: 9765 Total Submissions: 32502 My Submissions
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.

题目链接

这是我的代码：

主要的注意点是要考虑 root = NULL的情况

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool hasPathSum(TreeNode *root, int sum) {        if (root == NULL) return false;        if (root->left == NULL and root->right == NULL)            return root->val == sum ? true : false;        if (root->left != NULL and hasPathSum(root->left, sum-(root->val)))            return true;        if (root->right != NULL and hasPathSum(root->right, sum-(root->val)))            return true;        return false;    }};

这是我的Python代码：

class Solution:    # @param root, a tree node    # @param sum, an integer    # @return a boolean    def hasPathSum(self, root, sum):        if root is None:            return False        if root.left is None and root.right is None:            return True if root.val == sum else False        if root.left != None and self.hasPathSum(root.left, sum-(root.val)):            return True        if root.right != None and self.hasPathSum(root.right, sum-(root.val)):            return True        return False

Python的速度大约是C++的1/5到1/4