Given an integer 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
Solution:
this is a little difficult than the "96 Unique Binary Search Trees I"
it requires to get all the unique binary search tree stored.
1. 1st try to use iterative way
define dpT[i] indicate the treenodes stored in last all rooted as i
dpT[i]
dpT[0] = []
dpT[1] = [TreeNode(1)]
dp[2] = dp[0]*dp[1] + dp[1]*dp[0]
for e in dpT[1]
t2 = TreeNode(2)
t2.right = e
dpT[2].append(t2)
for e in dpT[1]
t1 = TreeNode(1)
t1.left = e
dpT[2].append(t1)
dp[3] = dp[0]*dp[2] + dp[1]*dp[1] + dp[2]*dp[0]
if i == 0:
for e in dpT[2] //k-i-1
root = TreeNode(1) // 1
root.right = e
dpT[3].append(root） //k
else if i == 2: //k-1
for e in dpT[2] //i
root = TreeNode(3) // k
root.left = e
dpT[3].append(root) //k
else:
for e1 in dpT[1]: //i
for e2 in dpT[1]: //k-i-1
root = TreeNode(2] // i e.g. 2 is the second element important
root.left = e1
root.right = e2
dpT[3].append(root) //k
2. use recursive way
leftsubtree (1, k-1)
root (k)
rightsubtree(k+1, n)
1 def generateTreeHelper(left, right): 2 lst = [] 3 if left > right: 4 lst.append(None) 5 return lst 6 7 for i in range(left, right+1, 1): 8 lSubTree = generateTreeHelper(left, i-1) 9 rSubTree = generateTreeHelper(i+1, right) 10 for j in range(0, len(lSubTree)): 11 for k in range(0, len(rSubTree)): 12 root = TreeNode(i) 13 root.left = lSubTree[j] 14 root.right = rSubTree[k] 15 lst.append(root) 16 return lst 17 if n == 0: 18 return [] 19 return generateTreeHelper(1, n)