split-bst.py

"""

LC 776 - Split BST

Problem Description
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.

"""

# V0
# IDEA : BST properties (left < root < right) + recursion
# https://blog.csdn.net/magicbean2/article/details/79679927
# https://www.itdaan.com/tw/d58594b92742689b5769f9827365e8b4
### STEPS
#  -> 1) check whether root.val > or < V
#     -> if root.val > V : 
#           - NO NEED TO MODIFY ALL RIGHT SUB TREE
#           - BUT NEED TO re-connect nodes in LEFT SUB TREE WHICH IS BIGGER THAN V (root.left = right)
#     -> if root.val < V : 
#           - NO NEED TO MODIFY ALL LEFT SUB TREE
#           - BUT NEED TO re-connect nodes in RIGHT SUB TREE WHICH IS SMALLER THAN V (root.right = left)
# -> 2) return result
class Solution(object):
    def splitBST(self, root, V):
        if not root: return [None, None]
        ### NOTE : if root.val <= V
        if root.val > V:
            left, right = self.splitBST(root.left, V)
            root.left = right
            return [left, root]
        ### NOTE : if root.val > V
        else:
            left, right = self.splitBST(root.right, V)
            root.right = left
            return [root, right]

# V0'
# IDEA : BST properties (left < root < right) + recursion
class Solution(object):
    def splitBST(self, root, V):
        if not root:
            return None, None
        ### NOTE : if root.val <= V
        elif root.val <= V:
            result = self.splitBST(root.right, V)
            root.right = result[0]
            return root, result[1]
        ### NOTE : if root.val > V
        else:
            result = self.splitBST(root.left, V)
            root.left = result[1]
            return result[0], root

# V0'
class Solution(object):
    def splitBST(self, root, V):
        if not root:
            return None, None
        elif root.val <= V:
            result = self.splitBST(root.right, V)
            root.right = result[0]
            return root, result[1]
        elif root.val > V:
            result = self.splitBST(root.left, V)
            root.left = result[1]
            return result[0], root

# V0''
class Solution(object):
    def splitBST(self, root, V):
        if not root: return [None, None]
        if root.val > V:
            left, right = self.splitBST(root.left, V)
            root.left = right
            return [left, root]
        left, right = self.splitBST(root.right, V)
        root.right = left
        return [root, right]

# V1 
# http://bookshadow.com/weblog/2018/02/04/leetcode-split-bst/
# https://blog.csdn.net/magicbean2/article/details/79679927
# https://www.itdaan.com/tw/d58594b92742689b5769f9827365e8b4
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution(object):
    def splitBST(self, root, V):
        """
        :type root: TreeNode
        :type V: int
        :rtype: List[TreeNode]
        """
        if not root: return [None, None]
        if root.val > V:
            left, right = self.splitBST(root.left, V)
            root.left = right
            return [left, root]
        left, right = self.splitBST(root.right, V)
        root.right = left
        return [root, right]

### Test case : dev 

# V1'
# https://blog.csdn.net/magicbean2/article/details/79679927
# JAVA
# /**
#  * Definition for a binary tree node.
#  * struct TreeNode {
#  *     int val;
#  *     TreeNode *left;
#  *     TreeNode *right;
#  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
#  * };
#  */
# class Solution {
# public:
#     vector<TreeNode*> splitBST(TreeNode* root, int V) {
#         vector<TreeNode *> res(2, NULL);
#         if(root == NULL) {
#             return res;
#         }
#         if(root->val > V) {         // the right child is retained
#             res[1] = root;
#             auto res1 = splitBST(root->left, V);
#             root->left = res1[1];
#             res[0]=res1[0];
#         }
#         else {                      // the left child is retained
#             res[0] = root;
#             auto res1 = splitBST(root->right, V);
#             root->right = res1[0];
#             res[1] = res1[1];
#         }
#         return res;
#     }
# };

# V1''
# https://www.itdaan.com/tw/d58594b92742689b5769f9827365e8b4
# C++
# class Solution {
# public:
#     vector<TreeNode*> splitBST(TreeNode* root, int V) {
#         vector<TreeNode*> res{NULL, NULL};
#         if (!root) return res;
#         if (root->val <= V) {
#             res = splitBST(root->right, V);
#             root->right = res[0];
#             res[0] = root;
#         } else {
#             res = splitBST(root->left, V);
#             root->left = res[1];
#             res[1] = root;
#         }
#         return res;
#     }
# };

# V1'''
# https://www.acwing.com/solution/LeetCode/content/208/
# IDEA : JAVA
# class Solution {
#     public TreeNode[] splitBST(TreeNode root, int V) {
#         TreeNode[] ans = new TreeNode[2];
#         return dfs(root, V);
#     }

#     public TreeNode[] dfs(TreeNode node, int v) {
#         TreeNode[] ans = new TreeNode[2];
#         if(node==null) return ans;

#         if(node.val>v) {
#             ans[1] = node;
#             TreeNode left = node.left;
#             node.left = null;
#             TreeNode[] nodes = dfs(left, v);
#             node.left = nodes[1];
#             ans[0] = nodes[0];
#         }else {
#             ans[0] = node;
#             TreeNode right = node.right;
#             node.right = null;
#             TreeNode[] nodes = dfs(right, v);
#             node.right = nodes[0];
#             ans[1]=nodes[1];
#         }
#         return ans;
#     }
#
# }

# V2 
# Time:  O(n)
# Space: O(h)
class Solution(object):
    def splitBST(self, root, V):
        """
        :type root: TreeNode
        :type V: int
        :rtype: List[TreeNode]
        """
        if not root:
            return None, None
        elif root.val <= V:
            result = self.splitBST(root.right, V)
            root.right = result[0]
            return root, result[1]
        else:
            result = self.splitBST(root.left, V)
            root.left = result[1]
            return result[0], root