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  • 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.


    题解:
    这道题比1难的就是不是返回个数,而是返回全部结果。从http://blog.csdn.net/sinat_24520925/article/details/45562273可知可行的二叉查找树的数量时对应的卡特兰数。由于左右子数也是一棵树,仅仅是树节点的值不同,所以我们用递归的方法求全部树的结构。分成左子树以及右子树,将1~n先分成两部分,之后将这两部分递归所组成子树挂在根节点下就可以,代码例如以下:
    /**
     * 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*>subTree(int start,int end)
        {
            vector<TreeNode*> res;
            if(start>end)
            {
                res.push_back(NULL);
                return res;
            }
            for(int k=start;k<=end;k++)
            {
                vector<TreeNode*> left=subTree(start,k-1);
                vector<TreeNode*> right=subTree(k+1,end);
                for(int i=0;i<left.size();i++)
                {
                    for(int j=0;j<right.size();j++)
                    {
                        TreeNode* node=new TreeNode(k);
                        node->left=left[i];
                        node->right=right[j];
                        res.push_back(node);
                    }
                }
            }
            return res;
            
        }
        vector<TreeNode*> generateTrees(int n) {
            vector<TreeNode*> res;
            res=subTree(1,n);
            return res;
        }
    };


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  • 原文地址:https://www.cnblogs.com/yfceshi/p/6820234.html
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