《leetCode》：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

思路

找到一个数作为根结点，剩余的数分别划入左子树或者右子树。

result of BSTs with k being the root

=root(k)+ result of BSTs of (1~k-1) numbers as leftSubTree +result of BSTs of (k+1~n) numbers as rightSubTree

实现代码如下：

/**
* 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<1){
return new ArrayList<TreeNode>();
}
return generateTrees(1,n);
}
//函数功能：树节点中的value由start~end构成的二叉搜索树的所有结果
private List<TreeNode> generateTrees(int start, int end) {
List<TreeNode> res=new ArrayList<TreeNode>();
if(end<start){
res.add(null);
return res;
}
for(int i=start;i<=end;i++){
//产生以i为根节点，start~i-1构成左子树，i+1~end构成右子树
List<TreeNode> leftSubTreeList=generateTrees(start,i-1);
List<TreeNode> rightSubTreeList=generateTrees(i+1,end);
//遍历并由根节点组合左子树和右子树构成最后的结果
for(TreeNode leftSubTree:leftSubTreeList){
for(TreeNode rightSubTree:rightSubTreeList){
TreeNode root=new TreeNode(i);
root.left=leftSubTree;
root.right=rightSubTree;
res.add(root);
}
}
}
return res;
}
}

从这两题可以看出，递归确实是一个难点。还是不能够理解其中的精髓。