100. Same Tree同样的树
[抄题]:
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
[暴力解法]:
时间分析:
空间分析:
[思维问题]:
基础弱到没有recursion的概念
[一句话思路]:
recursion就是嵌套
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- 从正反两方面想,把所有情况都想到:p,q val相不相等,p,q不空、一个空、2个空
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
要有默认情况
[复杂度]:Time complexity: O(n) Space complexity: O(n)
所有的点走一遍,时间复杂度就是n
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[关键模板化代码]:
return ((p.val == q.val) && (isSameTree(p.left, q.left)) && (isSameTree(p.right, q.right)));
[其他解法]:
非递归,用stack,很麻烦 属于没事找事
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public boolean isSameTree(TreeNode p, TreeNode q) { //both are null if (p == null && q == null) return true; //just one null if (p == null || q == null) return false; //p.val == q.val, recursion if (p.val == q.val) return (isSameTree(p.left, q.left)) && (isSameTree(p.right, q.right)); //p.val != q.val return false;//default } }
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public boolean isSameTree(TreeNode p, TreeNode q) { if (p == null && q == null) return true; if (p == null && q != null) return false; if (p != null && q == null) return false; return ((p.val == q.val) && (isSameTree(p.left, q.left)) && (isSameTree(p.right, q.right))); } }