【leetcode】112. Path Sum（Python & C++）

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112. Path Sum

题目链接

112.1 题目描述：

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.

112.2 解题思路：

1. 思路一：递归，首先判断节点是否为空，若为空，则直接返回false。如果节点恰好是叶子节点，且该节点的值恰好为sum，则返回true，否则，则进入递归，分别判断该节点的左右子树，且此时的sum变为sum-此节点的值。因为只有有一条路存在即可，所以是或操作。

112.3 C++代码：

1、思路一代码（12ms）：

class Solution103 {public:    bool hasPathSum(TreeNode* root, int sum) {        if (root == NULL)            return false;        if (root->left==NULL && root->right==NULL && root->val == sum)            return true;        else if (root->left == NULL && root->right == NULL && root->val != sum)            return false;        else            return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);    }};

112.4 Python代码：

1、思路二代码（65ms）：

class Solution(object):    def hasPathSum(self, root, sum):        """        :type root: TreeNode        :type sum: int        :rtype: bool        """        if root==None:            return False        if root.left==None and root.right==None and root.val==sum:            return True        elif root.left==None and root.right==None and root.val!=sum:            return False        else:            return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right,sum-root.val)

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