Problem
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.
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
Python 实现
'''
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.
'''
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def hasPathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: bool
"""
if root == None:
return False
leftValue = sum - root.val
if root.left == None and root.right == None and leftValue == 0:
return True
return self.hasPathSum(root.left, leftValue) or self.hasPathSum(root.right, leftValue)
分析
还是递归的思路。子树有满足减去当前节点值的和即可。