题目描述

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.

例子

例子 1

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

例子 2

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

Constraints

• The number of nodes in the tree in the range [1, 10^4].
• 0 <= Node.val <= 10^4
• The value of each node in the tree is unique.
• root is guaranteed to be a valid binary search tree.
• 0 <= low <= high <= 10^4

解题思路

这道题目借助二叉搜索树的结构可以比较简单的梳理出逻辑：

• 如果当前 root 为空，直接返回
• 如果当前 root 比下界还小，左子树全部丢弃，对右子树进行递归处理之后返回右子树的根节点即可
• 同理，如果当前 root 比上界还大，右子树全部丢弃，对左子树进行递归处理之后返回左子树的根节点即可
• 如果当前 root 在合法范围内，则对左右子树各自进行迭代操作，然后将处理后的左右子树根节点接上即可

class Solution {
public:
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        if (!root) {
            return nullptr;
        }
        if (root->val < low)
        {
            return trimBST(root->right, low, high);
        }
        else 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;
    }
};

• 时间复杂度: O(n)
• 空间复杂度: O(1)