95 Unique Binary Search Trees II

95. Unique Binary Search Trees II

1. Question

Given an integer n, generate all structurally uniqueBST'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                 32. Implementation

(1) Recursion

思路:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<TreeNode> generateTrees(int n) {
        if (n <= 0) {
            return new ArrayList<TreeNode>();
        }
        return generateTreesByDFS(1, n);
    }

    public List<TreeNode> generateTreesByDFS(int start, int end) {
        List<TreeNode> res = new ArrayList();

        if (start > end) {
            res.add(null);
            return res;
        }

        if (start == end) {
            res.add(new TreeNode(start));
            return res;
        }

        List<TreeNode> leftSubTrees = null;
        List<TreeNode> rightSubTrees = null;

        for (int i = start; i <= end; i++) {
            leftSubTrees = generateTreesByDFS(start, i - 1);
            rightSubTrees = generateTreesByDFS(i + 1, end);

            for (int j = 0; j < leftSubTrees.size(); j++) {
                for (int k = 0; k < rightSubTrees.size(); k++) {
                    TreeNode root = new TreeNode(i);
                    root.left = leftSubTrees.get(j);
                    root.right = rightSubTrees.get(k);
                    res.add(root);
                }
            }
        }
        return res;
    }
}

3. Time & Space Complexity

DP: 时间复杂度O(), 空间复杂度O(n)