[leetcode] 112. Path Sum @ python

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原题

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.

Note: A leaf is a node with no children.

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.

解法

递归, base case有3种情况: 1) root为空时, 返回False 2) root没有子树且root的值等于sum, 返回True 3) root没有子树且root的值不等于sum, 返回False. 然后再从root的左右子树中寻找结果.
Time: O(n) , n为节点的数量
Space: O(1)

代码

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

class Solution(object):
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if not root:
            return False
        if root.left is None and root.right is None and root.val == sum:
            return True
        if root.left is None and root.right is None and root.val != sum:
            return False
        return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)