95. Unique Binary Search Trees II

最新推荐文章于 2021-05-17 19:12:35 发布

SnailRunFast

最新推荐文章于 2021-05-17 19:12:35 发布

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本文探讨了如何生成所有可能的、结构上唯一的二叉搜索树，这些树存储从1到n的值。通过递归地创建左子树和右子树的组合，实现了这一目标。文章详细解释了算法的实现过程，包括使用后序遍历获取根节点，并展示了如何通过双重循环获取所有可能的组合。

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95. Unique Binary Search Trees II

Medium

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n.

Example:

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

Tips: Here I leverage postorder to traverse the binary tree, so I can get left and right node first, then create the root node. The difference between other recursive tree problem. The left and right children is a list instead of just a node, so I need to cope with the root node using double 'for' expression to get all the possible combination.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<TreeNode> generateTrees(int n) {
        if (n <= 0) {
            return new ArrayList<TreeNode>();
        }
        // node label from 1 to n
        return generateTrees(1, n);
    }
    
    
    private List<TreeNode> generateTrees(int left, int right) {
        if (left > right) {
            List<TreeNode> tResult = new ArrayList<>();
            tResult.add(null);
            return tResult;
        } else if (left == right) {
            return Arrays.asList(new TreeNode(left));
        } else {
            // left right root
            List<TreeNode> result = new ArrayList<>();
            for (int i = left; i <= right; i++) {
                List<TreeNode> leftChild = generateTrees(left, i - 1);
                List<TreeNode> rightChild = generateTrees(i + 1, right);
                for (TreeNode ln : leftChild) {
                    for (TreeNode rn: rightChild) {
                        TreeNode root = new TreeNode(i);
                        root.left = ln;
                        root.right = rn;
                        result.add(root);
                    }
                }
            }

            return result; 
        }
    }
}