小知识，大挑战！本文正在参与“程序员必备小知识”创作活动。
描述
Given the roots of two binary trees p and q, 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: p = [1,2,3], q = [1,2,3]
Output: true
Example 2:
Input: p = [1,2], q = [1,null,2]
Output: false
Example 3:
Input: p = [1,2,1], q = [1,1,2]
Output: false
Note:
The number of nodes in both trees is in the range [0, 100].
-10^4 <= Node.val <= 10^4
解析
根据题意，给出了两颗树 p 和 q ，要求判断这两个树是否完全一样。直接使用递归对两颗树的节点进行遍历，如果有一个节点不相同直接返回 False ，否则就返回 True 。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def isSameTree(self, p, q):
"""
:type p: TreeNode
:type q: TreeNode
:rtype: bool
"""
if not p and not q: return True
if not p or not q: return False
if p.val != q.val: return False
return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
运行结果
Runtime: 16 ms, faster than 82.88% of Python online submissions for Same Tree.
Memory Usage: 13.5 MB, less than 67.93% of Python online submissions for Same Tree.
解析
当然了还可以使用栈的结构来遍历所有的节点，逻辑和上面的类似。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def isSameTree(self, p, q):
"""
:type p: TreeNode
:type q: TreeNode
:rtype: bool
"""
def isEqual(p,q):
if not p and not q: return True
if not p or not q: return False
if p.val != q.val: return False
return True
stack = [[p,q]]
while stack:
p,q = stack.pop(0)
if not isEqual(p,q):
return False
if p and q:
stack.append([p.left, q.left])
stack.append([p.right, q.right])
return True
运行结果
Runtime: 16 ms, faster than 82.88% of Python online submissions for Same Tree.
Memory Usage: 13.4 MB, less than 89.64% of Python online submissions for Same Tree.
原题链接：leetcode.com/problems/sa…
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