[leetcode] path sum @ Python [recursion]

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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.

思路： 递归的暴力破解。

穷尽式的解决了所有子问题后，就放心的自我调用吧。

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class 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.val == sum and root.left == None and root.right == None: 
            return True
        return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)

[leetcode] path sum @ Python [recursion]

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原文地址：http://www.cnblogs.com/asrman/p/4021339.html