【剑指Offer】面试题28：对称的二叉树-优快云博客

本文介绍了一种判断二叉树是否对称的方法，包括递归和非递归两种实现方式。递归方法通过比较左右子树来判断对称性；非递归方法则利用两个队列依次存储左右子节点并进行比较。

import java.util.LinkedList;
import java.util.Queue;

/**
 * 面试题28：对称的二叉树
 * 题目：请实现一个函数，用来判断一棵二叉树是不是对称的。如果一棵二叉树和它的镜像一样，那么它是对称的。
 * @author
 * @create 2021-05-02 22:25
 */
public class Solution28 {
    public static void main(String[] args) {
        TreeNode root = new TreeNode(8);
        TreeNode node1 = new TreeNode(6);
        TreeNode node2 = new TreeNode(6);
        TreeNode node3 = new TreeNode(5);
        TreeNode node4 = new TreeNode(7);
        TreeNode node5 = new TreeNode(7);
        TreeNode node6 = new TreeNode(5);

        root.left = node1;
        root.right = node2;
        node1.left = node3;
        node1.right = node4;
        node2.left = node5;
        node2.right = node6;

        boolean symmetrical = isSymmetricalNoRecu(root);
        System.out.println(symmetrical);
    }

    /**
     * 方法一：递归实现
     * @param pRoot
     * @return
     */
    public static boolean isSymmetrical(TreeNode pRoot) {
        if (pRoot == null){
            return true;
        }
        return isSymmetrical(pRoot.left, pRoot.right);
    }

    public static boolean isSymmetrical(TreeNode pRoot1, TreeNode pRoot2){
        if (pRoot1 == null && pRoot2 == null){
            return true;
        }
        if (pRoot1 == null || pRoot2 == null){
            return false;
        }
        if (pRoot1.val != pRoot2.val){
            return false;
        }

        return isSymmetrical(pRoot1.left, pRoot2.right)
                && isSymmetrical(pRoot1.right, pRoot2.left);
    }

    /**
     * 方法二：非递归实现,使用队列
     * @param pRoot
     * @return
     */
    public static boolean isSymmetricalNoRecu(TreeNode pRoot) {
        if (pRoot == null){
            return true;
        }
        Queue<TreeNode> leftqueue = new LinkedList<>();
        Queue<TreeNode> rightqueue = new LinkedList<>();
        leftqueue.add(pRoot.left);
        rightqueue.add(pRoot.right);
        while (!leftqueue.isEmpty() && !rightqueue.isEmpty()){
            TreeNode leftNode = leftqueue.poll();
            TreeNode rightNode = rightqueue.poll();

            if (leftNode == null && rightNode == null){
                continue;
            }
            if (leftNode == null || rightNode == null){
                return false;
            }
            if (leftNode.val != rightNode.val){
                return false;
            }
            leftqueue.add(leftNode.left);
            rightqueue.add(rightNode.right);

            leftqueue.add(leftNode.right);
            rightqueue.add(rightNode.left);
        }
        return leftqueue.isEmpty() && rightqueue.isEmpty();
    }
}