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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.
这道题目要求构建出存储1...n的所有平衡二叉树，并且这些树是相互独立的，如果有M棵树，那么每个节点都要被new M次。我的想法就是遍历1~n，构建出每一个数字作为根节点时它的左子树和右子树，然后根节点加所有左右子树的组合就是我们要构建的BST。很明显这是递归的思想，代码如下：
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public List<TreeNode> list = new ArrayList(); 12 public List<TreeNode> generateTrees(int n) { 13 if(n<=0) return list; 14 list = trees(1, n); 15 return list; 16 } 17 18 public List<TreeNode> trees(int n, int m){ //构建BST 19 List<TreeNode> clist = new ArrayList(); 20 if(m<n){ 21 clist.add(null); 22 return clist; 23 } 24 if(m==n){ 25 TreeNode node = new TreeNode(n); 26 clist.add(node); 27 return clist; 28 } 29 for(int i = n; i<=m; i++){ 30 List<TreeNode> llist = trees(n, i-1);//构建所有的左子树 31 List<TreeNode> rlist = trees(i+1, m);//构建所有的右子树 32 for(TreeNode lnode:llist){ //生成以i为根节点的所有BST 33 for(TreeNode rnode:rlist){ 34 TreeNode root = new TreeNode(i); 35 root.left = copyTree(lnode); 36 root.right = copyTree(rnode); 37 clist.add(root); 38 } 39 } 40 } 41 return clist; 42 } 43 44 public TreeNode copyTree(TreeNode root){ //复制二叉树 45 if(root==null) return null; 46 else{ 47 TreeNode croot = new TreeNode(root.val); 48 croot.left = copyTree(root.left); 49 croot.right = copyTree(root.right); 50 return croot; 51 } 52 } 53 }
LeetCode OJ 95. Unique Binary Search Trees II
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原文地址：http://www.cnblogs.com/liujinhong/p/5456393.html
