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【leetcode刷题笔记】Unique Binary Search Trees II

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

题解:递归的枚举1~n的每个节点为根节点,然后递归的利用它左边的节点构造左子树,放在一个list里面;再利用它右边的节点构造右子树,也放在一个list里面;最终枚举两个list里面的左子树和右子树,构建一棵树。

代码如下:

 1 /** 2  * Definition for binary tree 3  * public class TreeNode { 4  *     int val; 5  *     TreeNode left; 6  *     TreeNode right; 7  *     TreeNode(int x) { val = x; left = null; right = null; } 8  * } 9  */10 public class Solution {11     private ArrayList<TreeNode> generate(int start,int end){12         ArrayList<TreeNode> answer = new ArrayList<TreeNode>();13         if(start > end){14             answer.add(null);15             return answer;16         }17         18         //for every node from start to right,make it as tree root and recursively build its left and right child tree19         for(int i = start; i <= end;i++){20             ArrayList<TreeNode> left = generate(start, i-1);21             ArrayList<TreeNode> right = generate(i+1, end);22             for(TreeNode l:left){23                 for(TreeNode r:right){24                     TreeNode root = new TreeNode(i);25                     root.left = l;26                     root.right = r;27                     answer.add(root);28                 }29             }30             31         }32         33         return answer;34         35     }36     public List<TreeNode> generateTrees(int n) {37         return generate(1,n);38     }39 }