LCR144翻转二叉树(力扣简单题,Java,递归+非递归)

发布于:2024-03-15 ⋅ 阅读:(70) ⋅ 点赞:(0)

目录

题目描述:

递归代码1:

递归代码2:

非递归代码(层次遍历):


题目描述:

        给定一棵二叉树的根节点 root,请左右翻转这棵二叉树,并返回其根节点。

        示例 1:

输入:root = [5,7,9,8,3,2,4]
输出:[5,9,7,4,2,3,8]

递归代码1:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if(root==null){
            return null;
        }
        TreeNode result=new TreeNode(root.val);
        getResult(root,result);
        return result;
    }
    public void getResult(TreeNode root,TreeNode copyRoot){
        if(root.right!=null){
            TreeNode node=new TreeNode(root.right.val);
            copyRoot.left=node;
            getResult(root.right,copyRoot.left);
        }
        if(root.left!=null){
            TreeNode node=new TreeNode(root.left.val);
            copyRoot.right=node;
            getResult(root.left,copyRoot.right);
        }
    }
}

递归代码2:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        fn(root);
        return root;
    }
    public void fn(TreeNode root){
        if(root==null){
            return ;
        }
        TreeNode t=root.left;
        root.left=root.right;
        root.right=t;

        fn(root.left);
        fn(root.right);
    }
}

非递归代码(层次遍历):

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if(root==null){
            return null;
        }
        Queue<TreeNode> queue=new LinkedList<>();
        queue.add(root);
        while(!queue.isEmpty()){
            int size=queue.size();
            for(int i=0;i<size;i++){
                TreeNode node=queue.remove();
                TreeNode t=node.right;
                node.right=node.left;
                node.left=t;
                if(node.right!=null){
                    queue.add(node.right);
                }
                if(node.left!=null){
                    queue.add(node.left);
                }
            }
        }
        return root;
    }
}