标签：
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.
The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.
Here‘s an example:
1
/ 2 3
/
4
5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".思路：n个数都是可以作为树的结点的，所以每层都选择一个结点后，然后再组合左右两个子树的情况。
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; left = null; right = null; }
* }
*/
public class Solution {
public List<TreeNode> generateTrees(int n) {
return dfs(0, n-1);
}
private List<TreeNode> dfs(int beg, int end) {
List<TreeNode> ans = new ArrayList<TreeNode>();
if (beg > end) {
ans.add(null);
return ans;
}
for (int i = beg; i <= end; i++) {
List<TreeNode> left = dfs(beg, i-1);
List<TreeNode> right = dfs(i+1, end);
for (int j = 0; j < left.size(); j++)
for (int k = 0; k < right.size(); k++) {
TreeNode tmp = new TreeNode(i+1);
ans.add(tmp);
tmp.left = left.get(j);
tmp.right = right.get(k);
}
}
return ans;
}
}
LeetCode Unique Binary Search Trees II
标签：
原文地址：http://blog.csdn.net/u011345136/article/details/45223149
