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.
解題思路：
從根到葉子結(jié)點唠亚，深度遍歷所有的路徑霹俺，直到找到一條符合條件的路徑焊傅。
Python實現(xiàn)：
# 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
if sum - root.val == 0 and root.left == None and root.right == None: # 如果累積到當前結(jié)點結(jié)果正好為0且是葉子結(jié)點
return True
else:
return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val) # 只需要找到一條路徑即可
