Leetcode 669. Trim a Binary Search Tree

题目链接：https://leetcode.cn/problems/trim-a-binary-search-tree/

方法一 迭代

1 方法思想

2 代码实现

/**
 * 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) {
        if (root == null) return root;
        while (root != null && (root.val < low || root.val > high)) {
            if (root.val < low) {
                root = root.right;
            } else{
                root = root.left;
            }        
        }

        TreeNode cur = root;
        while (cur != null && cur.left != null) {
            if (cur.left.val < low){
                cur.left = cur.left.right;
            }else {
                cur = cur.left;
            }
        }
        cur = root;
        while (cur != null &&  cur.right !=null){
            if (cur.right.val > high){
                cur.right = cur.right.left;
            }else {
                cur = cur.right;
            }
        }
        return root;
    }
}

3 复杂度分析

时间复杂度：
空间复杂度：

4 涉及到知识点

5 总结

方法二 递归

1 方法思想

2 代码实现

/**
 * 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) {

        if (root == null) return root;
        if (root.val > high) return trimBST(root.left, low, high);
        if (root.val < low) return trimBST(root.right, low, high);

        root.right = trimBST(root.right, low, high);
        root.left = trimBST(root.left, low, high);
        return root;
    }
}

3 复杂度分析

时间复杂度：
空间复杂度：

4 涉及到知识点

5 总结