LeetCode 剑指 Offer II 052. 展平二叉搜索树

文章目录

• LeetCode 剑指 Offer II 052. 展平二叉搜索树
• 题目描述
• 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
  • 2.知识点
• 总结
• 相同题目

题目描述

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

总结

相同题目

897. 递增顺序搜索树