LeetCode OJ

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：

/**
     * Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
     * Catalan number sequence
     * @param n
     * @return
     */
    public int numTrees(int n){
        if(n<=1)
            return 1;
        int[] f = new int[n+1];// the catalan
        f[0] = 1;
        f[1] = 1;
        int i=2;
        while(i<=n){
            int j = 0;
            while(j<=i-1){
                f[i] += f[j]*f[i-1-j];
                j++;
            }
            i++;
        }
        return f[n];
    }
    /**
     * Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.
     * @param n
     * @return
     */
     public ArrayList<TreeNode> generateTrees(int n) {
         
         if(n<=0)
         {
             ArrayList<TreeNode> r = new ArrayList<TreeNode>();
             r.add(null);
             return r;
         }
         return genTrees(1,n);
     }
     /**
      * 
      * @param start [
      * @param end ]
      * @return
      */
     private ArrayList<TreeNode> genTrees(int start, int end){
         if(start>end)
             return null;
         ArrayList<TreeNode> r = new ArrayList<TreeNode>();
         //only one element
         if(start == end)
         {
             r.add(new TreeNode(start));
             return r;
         }
         int i = start;
         while(i<=end){
             ArrayList<TreeNode> subL = genTrees(start,i-1);
             ArrayList<TreeNode> subR = genTrees(i+1,end);
             TreeNode root;
             //connect the left & right subtree
             if(subL!=null && subR!=null){
                 for(TreeNode left: subL){
                     for(TreeNode right:subR){
                         root = new TreeNode(i);
                         root.left = left;
                         root.right = right;
                         r.add(root);
                     }
                 }
             }
             if(subL == null && subR !=null){
                 for(TreeNode right: subR){
                     root = new TreeNode(i);
                     root.right = right;
                     r.add(root);
                 }
             }
             if(subR == null && subL !=null){
                 for(TreeNode left: subL){
                     root = new TreeNode(i);
                     root.left = left;
                     r.add(root);
                 }
             }
             i++;
         }
         return r;
     }