一文详解反转二叉树

1 前言 

LeetCode连接

根据二叉树的根节点root，反转这棵二叉树，并返回其根节点。

 

 2 思路

总体思想是采用层序遍历《Java 一文详解二叉树的层序遍历》

【第一步】交换root的左右子节点

 【第二步】交换节点7的左右子节点

 【第三步】交换节点2的左右子节点

package cn.msf.tree;

import javax.xml.namespace.QName;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.List;

import static cn.msf.tree.LevelTraversal.levelPrint;

/**
 * @author : msf
 * @date : 2022/12/6
 */
public class InvertTree {
    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        TreeNode node1 = new TreeNode(2);
        TreeNode node2 = new TreeNode(3);
        TreeNode node3 = new TreeNode(4);
        TreeNode node4 = new TreeNode(5);
        TreeNode node5 = new TreeNode(6);
        TreeNode node6 = new TreeNode(7);

        root.left = node1;
        root.right = node2;
        node1.left = node3;
        node1.right = node4;
        node2.left = node5;
        node2.right = node6;
        System.out.println("反转前 二叉树的层序遍历结果");
        List<ArrayList<Integer>> arrayLists = levelPrint(root);
        for (ArrayList<Integer> arrayList : arrayLists) {
            System.out.println(arrayList);
        }
        InvertTree tree = new InvertTree();
        root = tree.invertTree(root);
        System.out.println("反转后 二叉树的层序遍历结果");
        arrayLists  = levelPrint(root);
        for (ArrayList<Integer> arrayList : arrayLists) {
            System.out.println(arrayList);
        }
    }

    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        Deque<TreeNode> queue = new ArrayDeque<>();
        queue.addLast(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.removeFirst();
                if (node.left != null) {
                    queue.addLast(node.left);
                }
                if (node.right != null) {
                    queue.addLast(node.right);
                }
                if(node.left != null || node.right !=null) {
                    TreeNode temp = node.left;
                    node.left = node.right;
                    node.right = temp;
                }
            }
        }
        return root;
    }
}

上述代码执行结果：