【LeetCode每日一题】递增顺序搜索树

递增顺序搜索树

1、题目描述

给你一棵二叉搜索树，请你 按中序遍历 将其重新排列为一棵递增顺序搜索树，使树中最左边的节点成为树的根节点，并且每个节点没有左子节点，只有一个右子节点。

输入：root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
输出：[1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

输入：root = [5,1,7]
输出：[1,null,5,null,7]

2、算法描述

核心思想：
	题中给出了树是一个顺序搜索树，那么说明左边 > 中间 > 右边，我们可以采用中序遍历的方式进行求解

3、代码实现

（1）、树结构代码

package com.bean;

public class TreeNode {
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode() {
        
    }
    public TreeNode(int val) {
        this.val = val;
    }
    public TreeNode(int val, TreeNode left, TreeNode right) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

（2）、算法代码

package com.java;

import com.bean.TreeNode;

import java.util.*;

public class Day25_Solution {
    public TreeNode increasingBST(TreeNode root) {
        ArrayList list = new ArrayList();
        inOrder(root,list);
        TreeNode newRoot = new TreeNode(list.get(0));
        TreeNode currentNode = newRoot;

        for (int i=1;i list) {
        if (root != null) {
            inOrder(root.left,list);
            list.add(root.val);
            inOrder(root.right,list);
        }
    }

}

（3）、测试类代码

package com.test;

import com.bean.TreeNode;
import com.java.Day25_Solution;

public class Test25 {
    public static void main(String[] args) {
        Day25_Solution solution = new Day25_Solution();

        TreeNode node5 = new TreeNode(5);
        TreeNode node3 = new TreeNode(3);
        TreeNode node6 = new TreeNode(6);
        TreeNode node2 = new TreeNode(2);
        TreeNode node4 = new TreeNode(4);
        TreeNode node8 = new TreeNode(8);
        TreeNode node1 = new TreeNode(1);
        TreeNode node7 = new TreeNode(7);
        TreeNode node9 = new TreeNode(9);

        node5.left = node3;
        node5.right = node6;

        node3.left = node2;
        node3.right = node4;

        node6.left = null;
        node6.right = node8;

        node2.left = node1;

        node8.left = node7;
        node8.right = node9;

        TreeNode root = solution.increasingBST(node5);
        while (root != null) {
            System.out.println(root.val);
            root = root.right;
        }

    }
}