Given two binary trees, write a function to check if they are the same or not.
Two binary trees are considered the same if they are structurally identical and the nodes have the same value.
Example 1:
Input: 1 1 / \ / \ 2 3 2 3 [1,2,3], [1,2,3] Output: true
Example 2:
Input: 1 1 / \ 2 2 [1,2], [1,null,2] Output: false
Example 3:
Input: 1 1 / \ / \ 2 1 1 2 [1,2,1], [1,1,2] Output: false
LeetCode:链接
第一种:递归方法。
# 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 isSameTree(self, p, q):
"""
:type p: TreeNode
:type q: TreeNode
:rtype: bool
"""
if p == None and q == None:
return True
if p == None or q == None:
return False
return p.val == q.val and self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
第二种方法:非递归方法。
# 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 isSameTree(self, p, q):
"""
:type p: TreeNode
:type q: TreeNode
:rtype: bool
"""
'''像元组形式送进stack'''
stack = [(p, q)]
while stack:
p, q = stack.pop()
'''如果都为空 跳过这次循环'''
if p == None and q == None:
continue
if p == None or q == None:
return False
if p.val == q.val:
stack.append((p.left, q.left))
stack.append((p.right, q.right))
else:
return False
return True