[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]:

  1. 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] :

/**
* 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
}
}
/**
* 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)));
}
}

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