LeetCode | Unique Binary Search Trees II（构建二叉搜索树）

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}"

.

题目解析：

返回所有二叉搜索树的集合！

还是基于上一道题目的思想，依次选取1.....n个结点为根，然后递归遍历。但是由于要构造各种各样的树，因此要将返回的孩子跟结点集，一个一个的进行遍历

for(int i = begin;i <= end;i++){
vector<TreeNode *>leftTrees = GSubTrees(begin,i-1);
vector<TreeNode *>rightTrees = GSubTrees(i+1,end);
for(int j = 0;j < leftTrees.size();j++){
for(int k = 0;k < rightTrees.size();k++){
TreeNode *T = new TreeNode(i);//每次都要创建一个，不能共用！
T->left = leftTrees[j];
T->right = rightTrees[k];
tmp.push_back(T);
}
}
}

完整代码如下：

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<TreeNode *> generateTrees(int n) {

return GSubTrees(1,n);
}
vector<TreeNode *> GSubTrees(int begin,int end){
vector<TreeNode *> tmp;
if(begin>end){
tmp.push_back(NULL);
return tmp;
}
for(int i = begin;i <= end;i++){
vector<TreeNode *>leftTrees = GSubTrees(begin,i-1);
vector<TreeNode *>rightTrees = GSubTrees(i+1,end);
for(int j = 0;j < leftTrees.size();j++){
for(int k = 0;k < rightTrees.size();k++){
TreeNode *T = new TreeNode(i);//每次都要创建一个，不能共用！
T->left = leftTrees[j];
T->right = rightTrees[k];
tmp.push_back(T);
}
}
}
return tmp;
}
};