leetcode--UniqueBinarySearchTreesII

思路：利用遞歸的思想，要返回(1,n)之間的所有BST可以遍歷1->n，root作爲根節點，然後將(1->root-1)返回的所有節點作爲左子節點以及(root+1,n)返回的所有節點作爲右子節點拼接在root下，所有這些root放入數組返回。

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;

/**
 * Created by marsares on 15/6/18.
 */
public class UniqueBinarySearchTreesII {
    public List<TreeNode> generateTrees(int n) {
        return traversal(1,n);
    }
    public List<TreeNode>traversal(int start,int end){
        List<TreeNode>nodes=new ArrayList<TreeNode>();
        if(start>end){
            nodes.add(null);
            return nodes;
        }
        if(start==end){
            nodes.add(new TreeNode(start));
            return nodes;
        }
        for(int i=start;i<=end;i++){
            List<TreeNode>leftnodes=traversal(start,i-1);
            List<TreeNode>rightnodes=traversal(i+1,end);
            for(TreeNode leftnode:leftnodes){
                for(TreeNode rightnode:rightnodes){
                    TreeNode current=new TreeNode(i);
                    current.left=leftnode;
                    current.right=rightnode;
                    nodes.add(current);
                }
            }
        }
        return nodes;
    }
    public static void main(String[]args){
        UniqueBinarySearchTreesII ubst2=new UniqueBinarySearchTreesII();
        BinaryTreeSerialize bts=new BinaryTreeSerialize();
        for(TreeNode node:ubst2.generateTrees(4))System.out.println(bts.Serialize(node));
    }
}