1. 题目
2. 思路
(1) 层序遍历
- 对二叉树进行层序遍历，无论结点的子树是否为null都加入队列。
- 每遍历一层，先把所有结点加入集合，然后使用双指针法查看集合是否左右对称即可。
(2) 递归
- 每次比较时，若左右子树都为null，则返回true；若左右子树只有一个为null，则返回false；若左右子树都不为null，则比较其结点的值，并对左右子树进行递归。
- 递归时需要比较的是左子树的左子树和右子树的右子树，以及左子树的右子树和右子树的左子树。
(3) 迭代
3. 代码
import java.util.*;
public class Test {
public static void main(String[] args) {
}
}
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) {
if (root == null) {
return true;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<TreeNode> list = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
if (node != null) {
queue.offer(node.left);
queue.offer(node.right);
}
list.add(node);
}
int left = 0;
int right = list.size() - 1;
while (left < right) {
if (list.get(left) == null && list.get(right) == null || list.get(left) != null && list.get(right) != null && list.get(left).val == list.get(right).val) {
left++;
right--;
} else {
return false;
}
}
}
return true;
}
}
class Solution1 {
public boolean isSymmetric(TreeNode root) {
return check(root.left, root.right);
}
private boolean check(TreeNode left, TreeNode right) {
if (left == null && right == null) {
return true;
}
if (left == null || right == null) {
return false;
}
return left.val == right.val && check(left.left, right.right) && check(left.right, right.left);
}
}
class Solution2 {
public boolean isSymmetric(TreeNode root) {
if (root == null) {
return true;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root.left);
queue.offer(root.right);
while (!queue.isEmpty()) {
TreeNode node1 = queue.poll();
TreeNode node2 = queue.poll();
if (node1 == null && node2 == null) {
continue;
}
if ((node1 == null || node2 == null) || node1.val != node2.val) {
return false;
}
queue.offer(node1.left);
queue.offer(node2.right);
queue.offer(node1.right);
queue.offer(node2.left);
}
return true;
}
}