95. Unique Binary Search Trees II.md

95. Unique Binary Search Trees II

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leetcode-cn Daily Challenge on July 21th, 2020.

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Difficulty : Medium

Related Topics : Tree、Dynamic Programming

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Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n.

Example:

Input: 3
Output:
[
  [1,null,3,2],
  [3,2,null,1],
  [3,1,null,null,2],
  [2,1,3],
  [1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST's shown below:

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

Constraints:

• 0 <= n <= 8

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Solution

• mine
  • Java
    • DP Bottom-Up Runtime: 1 ms, faster than 95.74%, Memory Usage: 39.3 MB, less than 94.37% of Java online submissions

      // O(S)time O(S)space
      // S = dp[0].size() +  dp[1].size() + .. + dp[n].size() = 4^n / n^(1/2) 
      public List<TreeNode> generateTrees(int n) {
          if(n < 1){
              return new ArrayList<>();
          }
          List<TreeNode>[] dp = new List[n + 1];
          for (int i = 0; i <= n; i++) {
              dp[i] = new ArrayList<>();
          }
          dp[1] = new ArrayList<>(Arrays.asList(new TreeNode(1)));
          for (int i = 2; i <= n; i++) {
              for (int j = 0; j < i; j++) {
                  dp[i].addAll(getNodeList(dp, j, i - 1 - j));
              }
          }
          return dp[n];
      }
      
      private List<TreeNode> getNodeList(List<TreeNode>[] dp, int i, int j) {
          List<TreeNode> res = new ArrayList<>();
          int sizeI = i == 0 ? 1 : dp[i].size();
          int sizeJ = j == 0 ? 1 : dp[j].size();
          int size = sizeI * sizeJ;
          for (int k = 0; k < size; k++) {
              TreeNode root = new TreeNode(i + 1);
              root.left = i == 0 ? null : dp[i].get(k / sizeJ);
              root.right = j == 0 ? null : copyTree(dp[j].get(k % sizeJ), root.val);
              res.add(root);
          }
          return res;
      }
      
      private TreeNode copyTree(TreeNode treeNode, int sizeAdd) {
          if (treeNode == null) {
              return null;
          }
          TreeNode node = new TreeNode(treeNode.val + sizeAdd);
          node.left = copyTree(treeNode.left, sizeAdd);
          node.right = copyTree(treeNode.right, sizeAdd);
          return node;
      }
      

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• leetcode solution

• Runtime: 1 ms, faster than 95.74%, Memory Usage: 39.9 MB, less than 68.06% of Java online submissions

  //O(S)time  O(S)space
  // S = 4^n / n^(1/2) 
  public List<TreeNode> generateTrees(int n) {
      if (n == 0) {
          return new LinkedList<TreeNode>();
      }
      return generate_trees(1, n);
  }
  
  public LinkedList<TreeNode> generate_trees(int start, int end) {
      LinkedList<TreeNode> all_trees = new LinkedList<TreeNode>();
      if (start > end) {
          all_trees.add(null);
          return all_trees;
      }
  
      // pick up a root
      for (int i = start; i <= end; i++) {
          // all possible left subtrees if i is choosen to be a root
          LinkedList<TreeNode> left_trees = generate_trees(start, i - 1);
  
          // all possible right subtrees if i is choosen to be a root
          LinkedList<TreeNode> right_trees = generate_trees(i + 1, end);
  
          // connect left and right trees to the root i
          for (TreeNode l : left_trees) {
              for (TreeNode r : right_trees) {
                  TreeNode current_tree = new TreeNode(i);
                  current_tree.left = l;
                  current_tree.right = r;
                  all_trees.add(current_tree);
              }
          }
      }
      return all_trees;
  }
  

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