95.Unique Binary Search Trees II

Tags: Medium Dynamic Programming Tree

Links: https://leetcode.com/problems/unique-binary-search-trees-ii/

---

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

---

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<TreeNode*> generateTrees(int n) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        if (n == 0) return {};
        return solve(1, n);
    }

    vector<TreeNode*> solve(int start, int end)
    {
        if (start > end) return {nullptr};
        vector<TreeNode*> res;
        for (int i = start; i <= end; ++i) {
            auto left = solve(start, i - 1), right = solve(i + 1, end);
            for (auto a : left) {
                for (auto b : right) {
                    TreeNode *root = new TreeNode(i);
                    root -> left = a;
                    root -> right = b;
                    res.push_back(root);
                }
            }
        }
        return res;
    }
};

采用分治+递归的做法。

划分左右两个子数组，递归构造。刚开始时，将区间 [1, n] 当作一个整体，然后需要将其中的每个数字都当作根结点，其划分开了左右两个子区间，然后分别调用递归函数，会得到两个结点数组，接下来要做的就是从这两个数组中每次各取一个结点，当作当前根结点的左右子结点，然后将根结点加入结果 res 数组中即可