LeetCode #112 路径总和 树
题目描述
给定一个二叉树和一个目标和,判断该树中是否存在根节点到叶子节点的路径,这条路径上所有节点值相加等于目标和。
说明: 叶子节点是指没有子节点的节点。
示例:
给定如下二叉树,以及目标和 sum = 22
,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。
方法一:递归
# 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: TreeNode, sum: int) -> bool:
if not root : return False
sum -= root.val
if(not root.left and not root.right):
return sum == 0
return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)
- 时间复杂度:
- 空间复杂度:最坏 ,最好
方法二:迭代
# 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: TreeNode, sum: int) -> bool:
node_stack = [root]
sum_stack = [sum]
while node_stack:
node = node_stack.pop()
sum = sum_stack.pop()
if not node:
continue
sum -= node.val
if sum == 0 and not node.left and not node.right:
return True
node_stack.append(node.left)
node_stack.append(node.right)
sum_stack.append(sum)
sum_stack.append(sum)
return False