思路:利用递归的思想,要返回(1,n)之间的所有BST可以遍历1->n,root作为根节点,然后将(1->root-1)返回的所有节点作为左子节点以及(root+1,n)返回的所有节点作为右子节点拼接在root下,所有这些root放入数组返回。
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));
}
}