力扣0100——相同的树

相同的树

难度：简单

题目描述

给你两棵二叉树的根节点 p 和 q ，编写一个函数来检验这两棵树是否相同。

如果两个树在结构上相同，并且节点具有相同的值，则认为它们是相同的。

示例1

输入： p = [1,2,3], q = [1,2,3]
输出： true

示例2

输入： p = [1,2], q = [1,null,2]
输出： false

示例3

输入： p = [1,2,1], q = [1,1,2]
输出： false

题解

依据题意得，本体需要使用回溯法解题，基本思想就是遍历每个节点之后对比

想法代码

using System.Runtime.Intrinsics.Arm;public class TreeNode
{public int val;public TreeNode left;public TreeNode right;public TreeNode(int val = 0, TreeNode left = null, TreeNode right = null){this.val = val;this.left = left;this.right = right;}
}
class Solution
{public static void Main(String[] args){TreeNode p = new TreeNode(1){left = new TreeNode(2),right = new TreeNode(3)};TreeNode q = new TreeNode(1){left = new TreeNode(2),right = new TreeNode(1)};Solution solution = new Solution();Console.WriteLine(solution.IsSameTree(p,q));}public bool IsSameTree(TreeNode p, TreeNode q){if (p == null && q == null){return true;}if (p == null || q == null){return false;}return p.val == q.val && IsSameTree(p.left,q.left) && IsSameTree(p.right,q.right);}
}