2014
11-18

# LeetCode-Unique Binary Search Trees II[二叉树]

### Unique Binary Search Trees II

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.

OJ’s Binary Tree Serialization:

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}".

// LeetCode, Unique Binary Search Trees II
// 时间复杂度TODO，空间复杂度TODO
class Solution {
public:
vector<TreeNode *> generateTrees(int n) {
if (n == 0) return generate(1, 0);
return generate(1, n);
}
private:
vector<TreeNode *> generate(int start, int end) {
vector<TreeNode*> subTree;
if (start > end) {
subTree.push_back(nullptr);
return subTree;
}
for (int k = start; k <= end; k++) {
vector<TreeNode*> leftSubs = generate(start, k - 1);
vector<TreeNode*> rightSubs = generate(k + 1, end);
for (auto i : leftSubs) {
for (auto j : rightSubs) {
TreeNode *node = new TreeNode(k);
node->left = i;
node->right = j;
subTree.push_back(node);
}
}
}
return subTree;
}
};


Java代码:

/**
* 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> helper(int start, int n){
List<TreeNode> ans = new ArrayList<TreeNode>();
if(n <= 1){
if(n == 0) ans.add(null);
else ans.add(new TreeNode(start+1) );
return ans;
}
for(int i=1; i<=n; i++){
List<TreeNode> leftList = helper(start, i-1);
List<TreeNode> rightList = helper(start+i, n-i);
for(TreeNode left:leftList){
for(TreeNode right:rightList){
TreeNode root = new TreeNode(i+start); //
root.left = left;
root.right = right;
ans.add(root);
}
}
}
return ans;
}

public List<TreeNode> generateTrees(int n) {
return helper(0,n);
}
}

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2. I like your publish. It is great to see you verbalize from the coronary heart and clarity on this essential subject matter can be easily noticed.

3. A猴子认识的所有猴子和B猴子认识的所有猴子都能认识，这句话用《爱屋及乌》描述比较容易理解……

4. 换句话说，A[k/2-1]不可能大于两数组合并之后的第k小值，所以我们可以将其抛弃。
应该是，不可能小于合并后的第K小值吧