LeetCode 题解（156）: Unique Binary Search Trees II

题目：

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.

题解：
用每一个数做根，递归左右。
C++版：

class Solution {
public:
    vector<TreeNode*> generateTrees(int n) {
        return generateTree(1, n);
    }
    
    vector<TreeNode*> generateTree(int left, int right) {
        vector<TreeNode*> result;
        if(left > right) {
            result.push_back(NULL);
            return result;
        }
        for(int i = left; i <= right; i++) {
            vector<TreeNode*> leftRes = generateTree(left, i - 1);
            vector<TreeNode*> rightRes = generateTree(i + 1, right);
            for(int j = 0; j < leftRes.size(); j++) {
                for(int k = 0; k < rightRes.size(); k++) {
                    TreeNode* root = new TreeNode(i);
                    root->left = leftRes[j];
                    root->right = rightRes[k];
                    result.push_back(root);
                }
            }
        }
        return result;
    }
};

Java版：

public class Solution {
    public List<TreeNode> generateTrees(int n) {
        return generateTree(1, n);
    }
    
    public List<TreeNode> generateTree(int left, int right) {
        List<TreeNode> result = new ArrayList<TreeNode>();
        if(left > right) {
            result.add(null);
            return result;
        }
        for(int i = left; i <= right; i++) {
            List<TreeNode> leftRes = generateTree(left, i - 1);
            List<TreeNode> rightRes = generateTree(i + 1, right);
            for(int j = 0; j < leftRes.size(); j++) {
                for(int k = 0 ; k < rightRes.size(); k++) {
                    TreeNode root = new TreeNode(i);
                    root.left = leftRes.get(j);
                    root.right = rightRes.get(k);
                    result.add(root);
                }
            }
        }
        return result;
    }
}

Python版：

class Solution:
    # @param {integer} n
    # @return {TreeNode[]}
    def generateTrees(self, n):
        return self.generateTree(1, n)
        
    def generateTree(self, left, right):
        if left > right:
            return [None]
            
        result = []
        for i in range(left, right+1):
            leftRes = self.generateTree(left, i - 1)
            rightRes = self.generateTree(i + 1, right)
            for j in range(len(leftRes)):
                for k in range(len(rightRes)):
                    root = TreeNode(i)
                    root.left = leftRes[j]
                    root.right = rightRes[k]
                    result.append(root)
                    
        return result