GuilinDev

Lc0776

05 August 2008

776 Split BST

题目

Given a Binary Search Tree (BST) with root node 

1

root
, and a target value 
1

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 
1

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:

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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 
    1

    50
    .
  2. The BST is always valid and each node’s value is different.

分析

树的题目用递归。考虑当前root, 处理完后, 会返回两棵树，一个较大书，一颗较小树。

如果当前 root.val <= V, 那么所有的左子树root.left（包含root）都在较小树中,这时候将右子树split成较小部分和较大部分，并且root需要连接到右子树的较小部分（递归得到）; 如果root.val > V也是类似的情况。

时间复杂度O(logn)，如果是balanced BST. 最坏情况O(n)如果不平衡的话.

代码

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/**
 * 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};
        }
        
        TreeNode[] splitted;
        if(root.val<= V) {
            splitted = splitBST(root.right, V);
            root.right = splitted[0]; //连接右子树的较小部分
            splitted[0] = root;
        } else {
            splitted = splitBST(root.left, V);
            root.left = splitted[1]; //连接左子树中的较大部分
            splitted[1] = root;
        }
        
        return splitted;
    }    
}
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