[LeetCode] Unique Binary Search Trees II

标签：c++   leetcode   

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

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

解题思路：

分别以i(1<=i<=n)为根节点构造BST。采用递归方法。左子树为1...i-1构成的BST，右子树为i+1...n构成的BST，如此递归下去组装即可。

/**
 * 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) {
        return generateSubTrees(1, n);
    }
    
    //产生一棵子树
    vector<TreeNode*> generateSubTrees(int start, int end){
        vector<TreeNode*> result;

		if (start>end){
			result.push_back(NULL);
			return result;
		}
        
        for(int i=start; i<=end; i++){
            vector<TreeNode*> leftTree = generateSubTrees(start, i-1);
            vector<TreeNode*> rightTree = generateSubTrees(i+1, end);
            for(vector<TreeNode*>::iterator it1 = leftTree.begin(); it1!=leftTree.end(); it1++){
                for(vector<TreeNode*>::iterator it2 = rightTree.begin(); it2!=rightTree.end(); it2++){
                    TreeNode* root = new TreeNode(i);
                    root->left = *it1;
                    root->right = *it2;
                    result.push_back(root);
                }
            }
        }
        
        return result;
    }
};

[LeetCode] Unique Binary Search Trees II

标签：c++   leetcode   

原文地址：http://blog.csdn.net/kangrydotnet/article/details/47318233