LeetCode 95.不同的二叉搜索树 II

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95. 不同的二叉搜索树 II

95.Unique Binary Search Trees II (Medium)

Leetcode / 力扣

2020-7-26 10:17:45

题目描述

给定一个数字 n，要求生成所有值为 1…n 的二叉搜索树。

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Input: 3
Output:
[
  [1,null,3,2],
  [3,2,null,1],
  [3,1,null,null,2],
  [2,1,3],
  [1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST's shown below:

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

解题思路

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<TreeNode> generateTrees(int n) {
        if(n < 1) return new LinkedList<TreeNode>();
        return generateSubTrees(1, n);
    }

    private List<TreeNode> generateSubTrees(int s, int e){
        List<TreeNode> ret = new LinkedList<TreeNode>();
        if(s > e){
            ret.add(null);
            return ret;
        }
        // 枚举可行根节点
        for (int i = s; i <= e; ++i) {
            // 获得所有可行的左子树集合
            List<TreeNode> leftSubtrees = generateSubTrees(s, i - 1);
            // 获得所有可行的右子树集合
            List<TreeNode> rightSubtrees = generateSubTrees(i + 1, e);
            // 从左子树集合中选出一棵左子树，从右子树集合中选出一棵右子树，拼接到根节点上
            for (TreeNode left : leftSubtrees) {
                for (TreeNode right : rightSubtrees) {
                    TreeNode root = new TreeNode(i);
                    root.left = left;
                    root.right = right;
                    ret.add(root);
                }
            }
        }
        return ret;
    }
}