leetcode95 Unique Binary Search Trees II
题目：
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.
思路：
本题采取递归的思路。
传递的参数是开始数值(begin)和结束数值(end)。
当begin > end 时，返回空(注意不是null)；
当begin == end 时， 返回含有 new TreeNode(begin)结点的ArrayList；
当begin < end时，建立两个ArrayList来分别接收左右子树。
代码：
1 public List<TreeNode> generateTrees(int n){ 2 return generateTrees(1, n); 3 } 4 5 public List<TreeNode> generateTrees(int begin, int end){ 6 List<TreeNode> arr = new ArrayList<TreeNode>(); 7 if(begin > end){ 8 return arr; 9 } 10 if(begin == end){ 11 TreeNode ptr = new TreeNode(begin); 12 arr.add(ptr); 13 return arr; 14 } 15 for(int i = begin; i <= end; i++){ 16 List<TreeNode> left = new ArrayList<TreeNode>(); 17 List<TreeNode> right = new ArrayList<TreeNode>(); 18 left = generateTrees(begin, i-1); 19 right = generateTrees(i+1, end); 20 //注意判断left和right是否为空 21 //还有，要注意应该在最内层循环每次都新建根结点 22 if(left.size() == 0){ 23 if(right.size() == 0){ 24 TreeNode root = new TreeNode(i); 25 root.left = null; 26 root.right = null; 27 arr.add(root); 28 }else{ 29 for(TreeNode r: right){ 30 TreeNode ptr = new TreeNode(i); 31 ptr.left = null; 32 ptr.right = r; 33 arr.add(ptr); 34 } 35 } 36 }else{ 37 if(right.size() == 0){ 38 for(TreeNode l: left){ 39 TreeNode ptr = new TreeNode(i); 40 ptr.left = l; 41 ptr.right = null; 42 arr.add(ptr); 43 } 44 }else{ 45 for(TreeNode l: left){ 46 for(TreeNode r: right){ 47 TreeNode ptr = new TreeNode(i); 48 ptr.left = l; 49 ptr.right = r; 50 arr.add(ptr); 51 } 52 } 53 } 54 } 55 } 56 return arr; 57 }
