leetcode -- 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

 OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / 
 2   3
    /
   4
    
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

[解题思路]

划分成左右子树分别进行构造，当左右子树所有可能情况都构造完毕后，加上node即可

这里根节点可能情况为1,2,....,n

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; left = null; right = null; }
 * }
 */
public class Solution {
    public ArrayList<TreeNode> generateTrees(int n) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(n == 0){
            return generate(1, 0);
        }
        return generate(1, n);
    }
    
    public ArrayList<TreeNode> generate(int start, int end){
        ArrayList<TreeNode> subTree = new ArrayList<TreeNode>();
        if(start > end){
            subTree.add(null);
            return subTree;
        }
        
        for(int i = start; i <= end; i++){
            ArrayList<TreeNode> leftSubTree = generate(start, i - 1);
            ArrayList<TreeNode> rightSubTree = generate(i + 1, end);
            for(int j = 0; j < leftSubTree.size(); j++){
                for(int k = 0; k < rightSubTree.size(); k++){
                    TreeNode node = new TreeNode(i);
                    node.left = leftSubTree.get(j);
                    node.right = rightSubTree.get(k);
                    subTree.add(node);
                }
            }
        }
        return subTree;
    }
}