[LeetCode] 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]

Constraints:

• The number of nodes in the tree in the range [1, 104].
• 0 <= Node.val <= 104
• The value of each node in the tree is unique.
• root is guaranteed to be a valid binary search tree.
• 0 <= low <= high <= 104

修剪二叉搜索树。

给你二叉搜索树的根节点 root ，同时给定最小边界low 和最大边界 high。通过修剪二叉搜索树，使得所有节点的值在[low, high]中。修剪树不应该改变保留在树中的元素的相对结构（即，如果没有被移除，原有的父代子代关系都应当保留）。 可以证明，存在唯一的答案。

所以结果应当返回修剪好的二叉搜索树的新的根节点。注意，根节点可能会根据给定的边界发生改变。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/trim-a-binary-search-tree
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思路是递归。题意是给了low和high两个边界，请你去掉BST中节点值不在这个范围内的节点但是保持其他节点的相对关系。思路大概是如下几点

如果root为空，就返回root

如果root不为空且root.val < low则去看右子树

如果root不为空且root.val > high则去看左子树

递归地去检查root的左孩子和右孩子

时间O(n)

空间O(n)

Java实现

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

 

JavaScript实现

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} low
 * @param {number} high
 * @return {TreeNode}
 */
var trimBST = function (root, low, high) {
    // corner case
    if (root === null) {
        return null;
    }
    if (root.val < low) {
        return trimBST(root.right, low, high);
    }
    if (root.val > high) {
        return trimBST(root.left, low, high);
    }
    root.left = trimBST(root.left, low, high);
    root.right = trimBST(root.right, low, high);
    return root;
};

 

LeetCode 题目总结