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;
}
}

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