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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
算法思路：
分治法：求出左子树的list和右子树的list，然后循环两个子树，分别插入root的左右子树。
代码如下：
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 List<TreeNode> generateTrees(int n) { 12 return buildTree(1,n); 13 } 14 private List<TreeNode> buildTree(int begin,int end){ 15 List<TreeNode> res = new ArrayList<TreeNode>(); 16 if(begin > end){ 17 res.add(null); 18 return res; 19 } 20 List<TreeNode> left = new ArrayList<TreeNode>(); 21 List<TreeNode> right = new ArrayList<TreeNode>(); 22 for(int i = begin; i <= end; i++){ 23 left = buildTree(begin, i - 1); 24 right = buildTree(i + 1, end); 25 for(int j = 0; j < left.size(); j++){ 26 for(int k = 0; k < right.size(); k++){ 27 TreeNode node = new TreeNode(i); 28 node.left = left.get(j); 29 node.right = right.get(k); 30 res.add(node); 31 } 32 } 33 } 34 return res; 35 } 36 }
参考：
http://www.cnblogs.com/cheapcrook/archive/2013/01/29/2880903.html
[leetcode]Unique Binary Search Trees II,布布扣,bubuko.com
[leetcode]Unique Binary Search Trees II
标签：style blog http color io strong for ar
原文地址：http://www.cnblogs.com/huntfor/p/3915819.html
