LeetCode 776. Split BST

原题链接在这里：https://leetcode.com/problems/split-bst/

题目：

Given a Binary Search Tree (BST) with root node root, and a target value V, split the tree into two subtrees where one subtree has nodes that are all smaller or equal to the target value, while the other subtree has all nodes that are greater than the target value.  It's not necessarily the case that the tree contains a node with value V.

Additionally, most of the structure of the original tree should remain.  Formally, for any child C with parent P in the original tree, if they are both in the same subtree after the split, then node C should still have the parent P.

You should output the root TreeNode of both subtrees after splitting, in any order.

Example 1:

Input: root = [4,2,6,1,3,5,7], V = 2
Output: [[2,1],[4,3,6,null,null,5,7]]
Explanation:
Note that root, output[0], and output[1] are TreeNode objects, not arrays.

The given tree [4,2,6,1,3,5,7] is represented by the following diagram:

          4
        /   
      2      6
     /     / 
    1   3  5   7

while the diagrams for the outputs are:

          4
        /   
      3      6      and    2
            /            /
           5   7         1

Note:

1. The size of the BST will not exceed 50.
2. The BST is always valid and each node's value is different.

题解：

If current node value is larger than V, then perfor DFS on its left subtree.

The returned value is an array of 2 trees. First one is smaller than or equal to V, second is larger than V.

Then current node left child should be first one. Now current node including its left subtree are all smaller than or equal to V, use it as new first element of array, and existing second element to return.

Vice versa.

Time Complexity: O(h).

Space: O(h).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode[] splitBST(TreeNode root, int V) {
        if(root == null){
            return new TreeNode[]{null, null};
        }

        if(root.val > V){
            TreeNode [] left = splitBST(root.left, V);
            root.left = left[1];
            return new TreeNode[]{left[0], root};
        }else{
            TreeNode [] right = splitBST(root.right, V);
            root.right = right[0];
            return new TreeNode[]{root, right[1]};
        }
    }
}

类似Delete Node in a BST.