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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 confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ. 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}".
难度：98。没有做出来的一道题，很难，参考了其他人的想法，发现基本上只有一个版本，在纸上按照它的步骤推了几步终于看懂了。分析如下：
第一次遇到这种让你construct a BST的问题，而且结果并不是建立一个ArrayList<ArrayList<Integer>>型的，而是建立一个ArrayList<TreeNode>型的，这我就不知道怎么搞了，看了别人的才发现原来只要把几种方案的root放在最终的结果ArrayList<TreeNode>里面就好了，每个root有自己的left & right，但是不用放在这个ArrayList里面，只要指定好就可以了。Note that current node needed to be created for each possible sub-tree setting.
以上是本问题的难点和不熟悉的地方，本题的思路参考了http://blog.csdn.net/linhuanmars/article/details/24761437
choose one number i from n as root, pass all the numbers less than i to recursive call to construct the current root‘s left sub-tree, pass all the numbers greater than i to the recursive call to construct the current root‘s right sub-tree. Take n=5 for example, when we choose root node = 3, then put 1,2 to recursive call to construct root node‘s left sub-tree, and put 4,5 to recursive call to contract root‘s right subtree.
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; left = null; right = null; } 8 * } 9 */ 10 public class Solution { 11 public ArrayList<TreeNode> generateTrees(int n) { 12 return helper(1, n); 13 } 14 15 public ArrayList<TreeNode> helper(int left, int right){ 16 ArrayList<TreeNode> rs = new ArrayList<TreeNode>(); 17 if(left>right) rs.add(null); 18 else if(left==right) rs.add(new TreeNode(left)); 19 else{ 20 for(int i=left; i<=right; i++){ 21 ArrayList<TreeNode> leftNodeSet = helper(left, i-1); 22 ArrayList<TreeNode> rightNodeSet = helper(i+1, right); 23 24 for(int j=0; j<leftNodeSet.size(); j++){ 25 for(int k=0; k<rightNodeSet.size(); k++){ 26 TreeNode curr = new TreeNode(i); 27 curr.left = leftNodeSet.get(j); 28 curr.right = rightNodeSet.get(k); 29 rs.add(curr); 30 } 31 } 32 } 33 } 34 return rs; 35 } 36 }
Leetcode: Unique Binary Search Trees II
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原文地址：http://www.cnblogs.com/EdwardLiu/p/3960079.html
