95. 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.

链接： http://leetcode.com/problems/unique-binary-search-trees-ii/

题解：

题目提示用dp，但真用dp的话内存会不够。所以这次我们使用递归。一开始也没什么头绪，discussion里面有些大神写得很好。方法是 - 从1至n遍历数字时，每次把所有数字分为三部分，  当前数字，比当前数字小的部分，以及比当前数字大的部分， 使用新的list分别存储这两部分。从左部和右部分别按顺序取值，和当前i一起组合起来，成为当前i的一个解，当左右两部分遍历完毕以后，就得到了当前i的所有解。接着计算下一个i的解集。

Time Complexity - O(2ⁿ)， Space Complexity - O(2ⁿ)

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

递归构造得很巧妙，要多加练习。 

二刷:

跟一刷的方法一样，还是类似于mergesort的divide and conquer，先计算左右两边，然后用i创建root节点，接下来assign左子树和右子树并且把结果保存到res里。 200题以后也有一道和这个很类似，好像是burst balloon之类的。

Java:

Time Complexity - O(2ⁿ)， Space Complexity - O(2ⁿ)

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

三刷:

方法和二刷一样。要注意理解思路，如何构造辅助方法来进行递归。

Java:

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

Update: 每天进步一点点

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

Reference：

https://leetcode.com/discuss/22628/recursive-java-solution-make-binary-search-characteristic

https://leetcode.com/discuss/33003/java-recursive-solution-straight-forward

https://leetcode.com/discuss/10254/a-simple-recursive-solution

https://leetcode.com/discuss/3440/help-simplify-my-code-the-second-one                       <- Python

https://leetcode.com/discuss/9790/java-solution-with-dp

https://leetcode.com/discuss/81728/java-2ms-solution-beats-92%25