21_101. 对称二叉树
题目描述：
解题思路：
- 递归：这道题算是二刷了，知道递归是一个向左一个向右，结果还是没做出来。出手代码不会写，只能又看了题解。对于返回类型是Boolean的递归，最后让其子问题&&一下就行。
- 迭代：最开始将根节点入队两次，其实就可以看做是两颗树，把其中一颗逆序，然后两棵树成对成对的入队出队。
代码：
递归
//递归：
/**
* 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 isSymmetric(TreeNode root) {
return check(root, root);
}
public boolean check(TreeNode p, TreeNode q) {
if (p == null && q == null) {
return true;
}
if (p == null || q== null) {
return false;
}
return p.val == q.val && check(p.left, q.right) && check(p.right, q.left);
}
}
迭代
//迭代：
/**
* 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 isSymmetric(TreeNode root) {
Queue queue = new LinkedList<>();
queue.offer(root);
queue.offer(root);
while (!queue.isEmpty()) {
TreeNode u = queue.poll();
TreeNode v = queue.poll();
if (u == null && v == null) {
continue;
}
if ((u == null || v == null) || (u.val != v.val)) {
return false;
}
queue.offer(u.left);
queue.offer(v.right);
queue.offer(u.right);
queue.offer(v.left);
}
return true;
}
}