[LeetCode] 776. Split BST

baiy 发布于2019-08-16 13:45 / 582人阅读

Problem

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:

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

Solution

class Solution {
    public TreeNode[] splitBST(TreeNode root, int V) {
        TreeNode[] res = new TreeNode[2];
        if (root == null) return res;
        if (root.val <= V) {
            res[0] = root;
            TreeNode[] rightRes = splitBST(root.right, V);
            root.right = rightRes[0];
            res[1] = rightRes[1];
        } else {
            res[1] = root;
            TreeNode[] leftRes = splitBST(root.left, V);
            root.left = leftRes[1];
            res[0] = leftRes[0];
        }
        return res;
    }
}

GPU云服务器云服务器 linux 776 BST LeetCode python bst

文章版权归作者所有，未经允许请勿转载,若此文章存在违规行为，您可以联系管理员删除。

转载请注明本文地址：https://www.ucloud.cn/yun/72287.html

leetcode449. Serialize and Deserialize BST

摘要：题目要求将二叉搜索树序列化和反序列化，序列化是指将树用字符串的形式表示，反序列化是指将字符串形式的树还原成原来的样子。假如二叉搜索树的节点较多，该算法将会占用大量的额外空间。题目要求 Serialization is the process of converting a data structure or object into a sequence of bits so that...

Honwhy 2019-08-16 16:58 评论0 收藏0
Serialize and Deserialize Binary Tree & BST

摘要：思路理论上说所有遍历的方法都可以。但是为了使和的过程都尽量最简单，是不错的选择。用作为分隔符，来表示。复杂度代码思路这道题和之前不同，一般的树变成了，而且要求是。还是可以用，还是需要分隔符，但是就不需要保存了。 297. Serialize and Deserialize Binary Tree Serialization is the process of converting a...

eccozhou 2019-08-14 17:11 评论0 收藏0
【LeetCode 二叉树专项】把二叉搜索树转换为累加树（538）

摘要：解法一中序遍历分析由于给定了二叉搜索树，因此首先考虑中序遍历，使用示例，我们先来分别看一下二叉搜索树和累加树中序遍历的结果二叉搜索树二叉累加树。这里还是使用示例，我们再来观察一下二叉搜索树和累加树中序遍历的结果二叉搜索树二叉累加树。 ...

xcold 2021-11-18 10:02 评论0 收藏0
leetcode 315 Count of Smaller Numbers After Self 以

摘要：题目意思就是要一个个的返回当前的最小值。所以解法自然就是。我们需要找出被打乱的点并返回正确结果。然后将两个不正确的点记录下来，最后回原来正确的值。如果是叶子节点，或者只有一个子树。思想来自于的代码实现。跳过总结请点这里：https://segmentfault.com/a/11... BST最明显的特点就是root.left.val < root.val < root.right.v...

inapt 2019-08-14 17:20 评论0 收藏0
[Leetcode] Kth Smallest Element in a BST 二叉搜索树第k小节

摘要：中序遍历复杂度时间空间思路因为左节点小于根节点小于右节点，二叉搜索树的一个特性就是中序遍历的结果就是树内节点从小到大顺序输出的结果。这里采用迭代形式，我们就可以在找到第小节点时马上退出。这样我们就可以用二叉树搜索的方法来解决这个问题了。 Kth Smallest Element in a BST Given a binary search tree, write a function...

Dean 2019-08-14 12:26 评论0 收藏0