100. Same Tree

Problem

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.

https://leetcode.com/problems/same-tree/

Example 1:

Input: p = [1,2,3], q = [1,2,3]
Output: true

Example 2:

Input: p = [1,2,3], q = [1,2,3]
Output: true

Example 3:

Input: p = [1,2,1], q = [1,1,2]
Output: false

Constraints:

The number of nodes in both trees is in the range [0, 100].
-10⁴ <= Node.val <= 10⁴

Test Cases

python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:

solution_test.py下载完整源码
py
import pytest

import os
import sys
sys.path.append(os.path.dirname(os.path.dirname(__file__)))
from _utils.binary_tree import build_tree
from solution import Solution

null = None


@pytest.mark.parametrize('p, q, expected', [
    ([1,2,3], [1,2,3], True),
    ([1,2], [1,null,2], False),
    ([1,2,1], [1,1,2], False),
])
class Test:
    def test_solution(self, p, q, expected):
        sol = Solution()
        assert sol.isSameTree(build_tree(p), build_tree(q)) == expected

Thoughts

在 572. Subtree of Another Tree 中包含了。

按前序（pre-order，NLR）同步遍历两棵二叉树。对于当前节点，如果一边有而另一边缺失，或者两边节点的值不同，直接返回 false。

Code

solution.py下载完整源码
py
from typing import Optional


# Definition for a binary tree node.
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


class Solution:
    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
        stack = []
        while p or q or stack:
            if (not not p) != (not not q):
                return False
            elif p:
                if p.val != q.val:
                    return False
                stack.append((p, q))
                p = p.left
                q = q.left
            elif stack:
                p, q = stack.pop()
                p = p.right
                q = q.right

        return True

#Problem

#Test Cases

#Thoughts

#Code

Problem

Test Cases

Thoughts

Code