Unique Binary Search Trees

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本文详细探讨了如何计算给定范围内独特二叉搜索树的数量，并通过DFS思想生成所有可能的独特二叉搜索树结构。

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Unique Binary Search Trees

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?

class Solution {
public:
    int numTrees(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(n<=0)
            return 0;
        if (n == 1)
            return 1;
        if (n==2)
            return 2;
            
        int array[n+1];
        array[0]=1;
        array[1]=1;
        array[2] = 2;
            
        for(int i = 3; i<=n; i++){
            array[i] = 0;
            for(int j = 0; j<i;j++){
                array[i] += array[j]*array[i-1-j];
            }
        }
        return array[n];
    }
};

Unique Binary Search Trees II

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

这一问利用DFS的思想

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<TreeNode *> result;
        if (n<0)
            return result;
            
        return generateTree(0,n-1);
    }
    
    vector<TreeNode *> generateTree(int start, int end){
        vector<TreeNode *> result;
        if (start > end){
            result.push_back(NULL);
            return result;
        }
        
        for (int i = start; i < end+1; i++){
            vector<TreeNode *> lefts = generateTree(start,i-1);
            vector<TreeNode *> rights = generateTree (i+1,end);
            
            for (int j = 0; j<lefts.size();j++)
                for (int k = 0; k<rights.size();k++ ){
                    TreeNode* root = new TreeNode(i+1);
                    root->left = lefts[j];
                    root->right = rights[k];
                    result.push_back(root);
                }
                
        }
        return result;
    }
};