164题 Unique Binary Search Trees II

最新推荐文章于 2025-12-11 22:46:37 发布

原创最新推荐文章于 2025-12-11 22:46:37 发布 · 66 阅读

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这篇博客介绍了如何使用递归算法生成从1到n的所有唯一二叉搜索树，通过实例展示了如何构造TreeNode类，并实现generateTrees方法。重点在于理解如何利用左右子树的范围构建不同形态的BST。

Unique Binary Search Trees II

Description
Given n, generate all structurally unique BST’s (binary search trees) that store values 1…n.

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @paramn n: An integer
     * @return: A list of root
     */
    public List<TreeNode> generateTrees(int n) {
        // write your code here
        return generateTrees(1, n);//从1作为root开始，到n作为root结束

    }
     private ArrayList<TreeNode> generateTrees(int left, int right){
          ArrayList<TreeNode> res = new ArrayList<TreeNode>();
          if (left > right){
              res.add(null);
              return res;
         }
         for (int i = left; i <= right; i++){
             ArrayList<TreeNode> lefts = generateTrees(left, i-1);//以i作为根节点，左子树由[1,i-1]构成
             ArrayList<TreeNode> rights = generateTrees(i+1, right);//右子树由[i+1, n]构成
             for (int j = 0; j < lefts.size(); j++){
                 for (int k = 0; k < rights.size(); k++){
                    TreeNode root = new TreeNode(i);
                     root.left = lefts.get(j);
                     root.right = rights.get(k);
                     res.add(root);//存储所有可能行
                 }
             }
         }
         return res;
     }
}