leetcode-Unique Binary Search Trees II

Unique Binary Search Trees II

 Total Accepted: 10486 Total Submissions: 39239My Submissions

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

Have you been asked this question in an interview? 


這道題是比較簡單的遞歸題，在編輯器上能容易寫地出來，但提交後有錯誤，根據錯誤信息大概能知道自己錯在哪，但這實際反映了自己的邏輯思維不夠清晰，自己也嘗試在紙上寫代碼，感覺就是很難考慮全面，要多練練。

要找出1-n數字的所有BST，我這裏用遞歸的方法，每個數字都可以爲根節點，然後比它小的數字建立左子樹，比它大的數字建立右子樹，再組合起來。要考慮清楚的是如何分配內存來存儲所有樹的所有節點。

/**
 * Definition for binary tree
 * 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) {
        return generateTrees(1, n);
    }
protected:
    vector<TreeNode *> generateTrees(int start, int end) {
        vector<TreeNode *> ret;
        if (start > end) {
            ret.push_back(NULL);
            return ret;
        }
        if (start == end) {
            ret.push_back(new TreeNode(start));
            return ret;
        }
        
        for (int i = start; i <= end; ++i) {
            vector<TreeNode *> leftTrees = generateTrees(start, i - 1);
            vector<TreeNode *> rightTrees = generateTrees(i + 1, end);
            
            for (int j = 0; j < leftTrees.size(); ++j) {
                for (int k = 0; k < rightTrees.size(); ++k) {
                    TreeNode *node = new TreeNode(i);
                    node->left = leftTrees[j];
                    node->right = rightTrees[k];
                    ret.push_back(node);
                }
            }
        }
        
        return ret;
    }
};