0669. Trim a Binary Search Tree (M)

Trim a Binary Search Tree (M)

题目

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, 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

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题意

将BST中所有值不在指定范围内的结点删去，同时不改变BST的基本结构。

思路

递归处理：如果当前结点值小于左端点，则删去当前结点和左子树，递归处理右子树；如果当前结点值大于右端点，则删去当前结点和右子树，递归处理左子树；否则保留当前结点，递归处理左右子树。

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代码实现

Java

class Solution {
    public TreeNode trimBST(TreeNode root, int low, int high) {
        if (root == null) {
            return null;
        }

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