leetcode解题: Unique Binary Search Trees II (95)

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Given an integer 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.

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1         3     3      2      1
 \       /     /      / \      \
  3     2     1      1   3      2
 /     /       \                 \
2     1         2                 3

解法1： Recursive, Divide & Conquer

用分治的思想考虑会比较容易，从1到n，如果选择了i，那么1到i-1所能组成的tree都是i的左子树，i+1到n都为右子树。对于每一颗左子树和右子树的组合，我们都能构建一个新的树。
用递归的方式很容易解决。
C++

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/**
 * Definition for a binary tree node.
 * 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) {
        vector<TreeNode*> res;
        if (n == 0) {
            return res;
        }
        return generate(1, n);
    }
    vector<TreeNode*> generate(int start, int end) {
        vector<TreeNode*> res;
        if (start > end) {
            res.push_back(NULL);
            return res;
        }
        for (int i = start; i<= end; ++i) {
            vector<TreeNode*> left = generate(start, i -1);
            vector<TreeNode*> right = generate(i + 1, end);
            for (TreeNode* l: left) {
                for (TreeNode* r: right) {
                    TreeNode* root = new TreeNode(i);
                    root->left = l;
                    root->right = r;
                    res.push_back(root);
                }
            }
        }
        return res;
    }
};

Java

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<TreeNode> generateTrees(int n) {
        List<TreeNode> res = new ArrayList<>();
        if (n <= 0) {
            return res;
        }
        return generateTrees(1, n);
    }
    private List<TreeNode> generateTrees(int start, int end) {
        List<TreeNode> res = new ArrayList<>();
        if (start > end) {
            return res;
        }
        if (start == end) {
            res.add(new TreeNode(start));
            return res;
        }
        for (int i = start; i <= end; i++) {
            List<TreeNode> left = generateTrees(start, i - 1);
            List<TreeNode> right = generateTrees(i + 1, end);
            if (left.isEmpty()) {
                for (TreeNode x : right) {
                    TreeNode root = new TreeNode(i);
                    root.right = x;
                    res.add(root);
                }
            } else if (right.isEmpty()) {
                for (TreeNode x : left) {
                    TreeNode root = new TreeNode(i);
                    root.left = x;
                    res.add(root);
                }
            } else {
                for (TreeNode l : left) {
                    for (TreeNode r : right) {
                        TreeNode root = new TreeNode(i);
                        root.left = l;
                        root.right = r;
                        res.add(root);
                    }
                }
            }
        }
        return res;
    }
}