LeetCode 剑指 Offer II 052. 展平二叉搜索树

LeetCode 剑指 Offer II 052. 展平二叉搜索树

题目描述

给你一棵二叉搜索树，请 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。
  示例 1：
  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]

LeetCode 剑指 Offer II 052. 展平二叉搜索树
提示：

    树中节点数的取值范围是 [1, 100]
    0 <= Node.val <= 1000

一、解题关键词

二、解题报告

1.思路分析

2.时间复杂度

3.代码示例

方法一

/**
 * Definition for a binary tree node.
 * public 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 TreeNode increasingBST(TreeNode root) {
        //递归展开 
        List<Integer> res = new ArrayList<>();
        inorder(root,res);

        TreeNode dummyNode = new TreeNode(-1);
        TreeNode currNode = dummyNode;
        for(int value : res){
            currNode.right = new TreeNode(value);
            currNode = currNode.right;
        }
        return dummyNode.right;
    }
    //中序遍历
    void inorder(TreeNode root,List<Integer> res){
        if(root == null) return;
        inorder(root.left,res);
        res.add(root.val);
        inorder(root.right,res);

    }
}

方法二

/**
 * Definition for a binary tree node.
 * public 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 TreeNode increasingBST(TreeNode root) {
        //递归展开 
        List<Integer> res = new ArrayList<>();
        inorder(root,res);

        TreeNode dummyNode = new TreeNode(-1);
        TreeNode currNode = dummyNode;
        for(int value : res){
            currNode.right = new TreeNode(value);
            currNode = currNode.right;
        }
        return dummyNode.right;
    }
    //中序遍历
    void inorder(TreeNode root,List<Integer> res){
        if(root == null) return;
        inorder(root.left,res);
        res.add(root.val);
        inorder(root.right,res);

    }
}

2.知识点

树的遍历 遍历之后重新组装数据

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总结

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