LeetCode 112 — Path Sum（C++ Java Python）

题目：http://oj.leetcode.com/problems/path-sum/

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.

题目翻译：

给定一个二叉树及总和，确定该树是否存在一条根至叶的路径，使得沿该路径的所有值加起来等于给定的总和。
例如：
给定如上二叉树，sum = 22，
返回true，因为存在一条根至叶的路径5->4->11->2，其总和为22。

分析：

        递归实现。

        如果节点为空则返回false；

        如果左右子节点均为空，则返回节点值是否等于sum；

        否则递归判断其左右子树，新sum为sum减去节点的值。

C++实现：

/** * 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 && root->right == NULL)    {    return sum == root->val;    }    return hasPathSum(root->left, sum - root->val)         || hasPathSum(root->right, sum - root->val);    }};

Java实现：

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean hasPathSum(TreeNode root, int sum) {        if (root == null) {return false;}if (root.left == null && root.right == null) {return root.val == sum;}return hasPathSum(root.left, sum - root.val)|| hasPathSum(root.right, sum - root.val);    }}

Python实现：

# Definition for a  binary tree node# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution:    # @param root, a tree node    # @param sum, an integer    # @return a boolean    def hasPathSum(self, root, sum):        if root == None:            return False                if root.left == None and root.right == None:            return root.val == sum                return self.hasPathSum(root.left, sum - root.val) \                or self.hasPathSum(root.right, sum - root.val)

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