花花酱 LeetCode 95. Unique Binary Search Trees II

zxi — Wed, 07 Mar 2018 17:20:29 +0000

Problem

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

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

Idea: Recursion

for i in 1..n: pick i as root,
left subtrees can be generated in the same way for n_l = 1 … i – 1,
right subtrees can be generated in the same way for n_r = i + 1, …, n
def gen(s, e):
return [tree(i, l, r) for l in gen(s, i – 1) for r in gen(i + 1, e) for i in range(s, e+1)

# of trees:

n = 0: 1
n = 1: 1
n = 2: 2
n = 3: 5
n = 4: 14
n = 5: 42
n = 6: 132
…
Trees(n) = Trees(0)*Trees(n-1) + Trees(1)*Trees(n-2) + … + Tress(n-1)*Trees(0)

Time complexity: O(3^n)

Space complexity: O(3^n)

C++

// Author: Huahua
// Running time: 16 ms
class Solution {
public:
  vector generateTrees(int n) {
    if (n == 0) return {};
    const auto& ans = generateTrees(1, n);
    cout << ans.size() << endl;
    return ans;
  }
private:
  vector generateTrees(int l, int r) {
    if (l > r) return { nullptr };
    vector ans;
    for (int i = l; i <= r; ++i)
      for (TreeNode* left : generateTrees(l, i - 1))
        for (TreeNode* right : generateTrees(i + 1, r)) {
          ans.push_back(new TreeNode(i));
          ans.back()->left = left;
          ans.back()->right = right;
        }
    return ans;
  }
};

Java

// Author: Huahua 4 ms
class Solution {
  public List generateTrees(int n) {
    if (n == 0) return new ArrayList();
    return generateTrees(1, n);
  }
  
  private List generateTrees(int l, int r) {
    List ans = new ArrayList<>();
    if (l > r) {      
      ans.add(null);
      return ans;
    }
    for (int i = l; i <= r; ++i)
      for (TreeNode left : generateTrees(l, i - 1))
        for (TreeNode right: generateTrees(i + 1, r)) {
          TreeNode root = new TreeNode(i);
          root.left = left;
          root.right = right;
          ans.add(root);
        }          
    return ans;
  }
}

Python 3

"""
Author: Huahua
Running time: 84 ms
"""
class Solution:
  def generateTrees(self, n):
    def newTree(x, l, r): 
      t = TreeNode(x)
      t.left = l
      t.right = r
      return t
    
    def gen(s, e):      
      return [newTree(i, l, r) for i in range(s, e + 1) for l in gen(s, i - 1) for r in gen(i + 1, e)] or [None]
        
    return gen(1, n) if n > 0 else []

Solution 2: DP

Java

// Author: Huahua, 3 ms
class Solution {
  public List generateTrees(int n) {
    if (n == 0) return new ArrayList();
    List[] dp = new List[n + 1];
    dp[0] = new ArrayList();
    dp[0].add(null);
    for (int i = 1; i <= n; ++i) {
      dp[i] = new ArrayList();
      for (int j = 0; j < i; ++j)
        for (TreeNode left : dp[j])
          for (TreeNode right: dp[i - j - 1]) {
            TreeNode root = new TreeNode(j + 1);
            root.left = left;
            root.right = clone(right, j + 1);
            dp[i].add(root);
        }          
      }
    return dp[n];
  }
  
  private TreeNode clone(TreeNode root, int delta) {
    if (root == null) return root;
    TreeNode node = new TreeNode(root.val + delta);
    node.left = clone(root.left, delta);
    node.right = clone(root.right, delta);
    return node;
  }
}

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花花酱 LeetCode 95. Unique Binary Search Trees II

Problem

Idea: Recursion

C++

Java

Python 3

Solution 2: DP

Java

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