LeetCode OJ - Unique Binary Search Tree 1 && 2

第一道题是Catalan数，主需要求f(n)，第二道题是用递归的方法，不断的组装这个棵树。

下面是两个AC代码：

 1 /**
 2      * Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
 3      * Catalan number sequence
 4      * @param n
 5      * @return
 6      */
 7     public int numTrees(int n){
 8         if(n<=1)
 9             return 1;
10         int[] f = new int[n+1];// the catalan
11         f[0] = 1;
12         f[1] = 1;
13         int i=2;
14         while(i<=n){
15             int j = 0;
16             while(j<=i-1){
17                 f[i] += f[j]*f[i-1-j];
18                 j++;
19             }
20             i++;
21         }
22         return f[n];
23     }
24     /**
25      * Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.
26      * @param n
27      * @return
28      */
29      public ArrayList<TreeNode> generateTrees(int n) {
30          
31          if(n<=0)
32          {
33              ArrayList<TreeNode> r = new ArrayList<TreeNode>();
34              r.add(null);
35              return r;
36          }
37          return genTrees(1,n);
38      }
39      /**
40       * 
41       * @param start [
42       * @param end ]
43       * @return
44       */
45      private ArrayList<TreeNode> genTrees(int start, int end){
46          if(start>end)
47              return null;
48          ArrayList<TreeNode> r = new ArrayList<TreeNode>();
49          //only one element
50          if(start == end)
51          {
52              r.add(new TreeNode(start));
53              return r;
54          }
55          int i = start;
56          while(i<=end){
57              ArrayList<TreeNode> subL = genTrees(start,i-1);
58              ArrayList<TreeNode> subR = genTrees(i+1,end);
59              TreeNode root;
60              //connect the left & right subtree
61              if(subL!=null && subR!=null){
62                  for(TreeNode left: subL){
63                      for(TreeNode right:subR){
64                          root = new TreeNode(i);
65                          root.left = left;
66                          root.right = right;
67                          r.add(root);
68                      }
69                  }
70              }
71              if(subL == null && subR !=null){
72                  for(TreeNode right: subR){
73                      root = new TreeNode(i);
74                      root.right = right;
75                      r.add(root);
76                  }
77              }
78              if(subR == null && subL !=null){
79                  for(TreeNode left: subL){
80                      root = new TreeNode(i);
81                      root.left = left;
82                      r.add(root);
83                  }
84              }
85              i++;
86          }
87          return r;
88      }

 

posted @ 2014-05-06 15:35  echoht  阅读(149)  评论(0)  编辑  收藏  举报