|
| 1 | +packageAlgorithms.tree; |
| 2 | + |
| 3 | +importjava.util.ArrayList; |
| 4 | +importjava.util.List; |
| 5 | +/* |
| 6 | + * Given n, generate all structurally unique BST's (binary search trees) that store values 1...n. |
| 7 | +
|
| 8 | +For example, |
| 9 | +Given n = 3, your program should return all 5 unique BST's shown below. |
| 10 | +
|
| 11 | + 1 3 3 2 1 |
| 12 | + \ / / / \ \ |
| 13 | + 3 2 1 1 3 2 |
| 14 | + / / \ \ |
| 15 | + 2 1 2 3 |
| 16 | +confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ. |
| 17 | + * */ |
| 18 | + |
| 19 | +/** |
| 20 | + * Definition for binary tree |
| 21 | + * public class TreeNode { |
| 22 | + * int val; |
| 23 | + * TreeNode left; |
| 24 | + * TreeNode right; |
| 25 | + * TreeNode(int x) { val = x; left = null; right = null; } |
| 26 | + * } |
| 27 | + */ |
| 28 | +publicclassGenerateTree2 { |
| 29 | +publicList<TreeNode>generateTrees(intn) { |
| 30 | +returngenerateTreesHelp(1,n); |
| 31 | + } |
| 32 | + |
| 33 | +/* |
| 34 | + 使用递归来完成，我们可以分解为2个步骤： |
| 35 | + 完成左子树，完成右子树。 |
| 36 | +
|
| 37 | + 如果说左子树有n种组合，右子树有m种组合，那最终的组合数就是n*m. 把这所有的组合组装起来即可 |
| 38 | + */ |
| 39 | +publicList<TreeNode>generateTreesHelp(intstart,intend) { |
| 40 | +ArrayList<TreeNode>ret =newArrayList<TreeNode>(); |
| 41 | + |
| 42 | +// null也是一种解，也需要把它加上去。这样在组装左右子树的时候，不会出现左边没有解的情况，或 |
| 43 | +// 是右边没有解的情况 |
| 44 | +if (start >end) { |
| 45 | +ret.add(null); |
| 46 | +returnret; |
| 47 | + } |
| 48 | + |
| 49 | +for (inti =start;i <=end;i++) { |
| 50 | +// 求出左右子树的所有的可能。 |
| 51 | +List<TreeNode>left =generateTreesHelp(start,i -1); |
| 52 | +List<TreeNode>right =generateTreesHelp(i +1,end); |
| 53 | + |
| 54 | +// 将左右子树的所有的可能性全部组装起来 |
| 55 | +for (TreeNodel:left) { |
| 56 | +for(TreeNoder:right) { |
| 57 | +// 先创建根节点 |
| 58 | +TreeNoderoot =newTreeNode(i); |
| 59 | +root.left =l; |
| 60 | +root.right =r; |
| 61 | + |
| 62 | +// 将组合出来的树加到结果集合中。 |
| 63 | +ret.add(root); |
| 64 | + } |
| 65 | + } |
| 66 | + } |
| 67 | + |
| 68 | +returnret; |
| 69 | + } |
| 70 | +} |