Leetcode 95: Unique Binary Search Trees II

https://leetcode.com/problems/unique-binary-search-trees-ii/

Given n, generate all structurally unique BST’s (binary search trees) that store values 1…n.

For example,
Given n = 3, your program should return all 5 unique BST’s shown below.

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

 

Code

# 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 copy(self, root, bias):
        if not root:
            return root
        
        new_root = TreeNode(root.val + bias)
        new_root.left = self.copy(root.left, bias)
        new_root.right = self.copy(root.right, bias)
        return new_root
        
    def generateTrees(self, n):
        """
        :type n: int
        :rtype: List[TreeNode]
        """
        if n == 0:
            return []
            
        ans = [[] for _ in xrange(n+1)]
        ans[0] = [None]
        
        for i in xrange(1, n+1):
            for j in xrange(1, i+1):    # j is the root value
                for lt in ans[j-1]:
                    for rt in ans[i-j]:
                        root = TreeNode(j)
                        root.left = self.copy(lt, 0)
                        root.right = self.copy(rt, j)
                        ans[i] += [root]
        return ans[-1]

 

Idea

To extend the idea of Leetcode 96, when you are given `n`, you can put `1~n` as roots. When you put `i` (`1<=i<=n`) as the root, the left tree is depending on the result of `i-1` and the right tree is depending on the result of `n-i`. For this problem, the right tree is not solely copy of the trees of `n-i`. Every node in the trees of `n-i` should be added a bias of `i`. Therefore, we create a recursive function `copy(self, root, bias)` to copy trees with biases.