[刷题防痴呆] 0669 - 修剪二叉搜索树 (Trim a B

题目地址

https://leetcode.com/problems/trim-a-binary-search-tree/

题目描述

669. Trim a Binary Search Tree

Given the root of a binary search tree and the lowest and highest boundaries as low and high, trim the tree so that all its elements lies in [low, high]. Trimming the tree should not change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a unique answer.

Return the root of the trimmed binary search tree. Note that the root may change depending on the given bounds.

 

Example 1:


Input: root = [1,0,2], low = 1, high = 2
Output: [1,null,2]
Example 2:


Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3
Output: [3,2,null,1]
Example 3:

Input: root = [1], low = 1, high = 2
Output: [1]
Example 4:

Input: root = [1,null,2], low = 1, high = 3
Output: [1,null,2]
Example 5:

Input: root = [1,null,2], low = 2, high = 4
Output: [2]

思路

• 对于原来的BST来说, 三种不同情况.
• 若根节点的值小于最小值，根节点需要修改, 把左节点全部裁掉, 根节点变为右子树节点, 递归调用右子树并返回右子树.
• 若根节点的值大于最大值，根节点需要修改, 把右节点全部裁掉, 根节点变为左子树节点, 递归调用左子树并返回左子树.
• 否则修剪左子树，右子树并返回根节点.

关键点

代码

• 语言支持：Java

/**
 * 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 trimBST(TreeNode root, int low, int high) {
        if (root == null) {
            return null;
        }
        if (root.val > high) {
            root = root.left;
            return trimBST(root, low, high);
        } else if (root.val < low) {
            root = root.right;
            return trimBST(root, low, high);
        } else {
            root.left = trimBST(root.left, low, high);
            root.right = trimBST(root.right, low, high);
        }

        return root;
    }
}