Leetcode 95 Unique Binary Search Trees II

triplebee

发布于 2018-01-12 14:59:38

4250

发布于 2018-01-12 14:59:38

Given an integer 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

求出所有二叉排序数的情况。

每一层枚举中间的点，将小于它的放在左子树，大于的放在右子树，没有元素的子树记得指向NULL。

/**
 * 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*> dfs(int l,int r)
    {
        if(l==r)
        { 
            TreeNode* p=new TreeNode(l);
            vector<TreeNode*> now(1,p);
            return now;
        }
        vector<TreeNode*> res;
        for(int mid=l;mid<=r;mid++)
        {
            vector<TreeNode*> left,right;
            if(l<=mid-1) 
                left=dfs(l,mid-1);
            else
                left.push_back(NULL);//开始忘记了
            if(mid+1<=r) 
                right=dfs(mid+1,r);
            else
                right.push_back(NULL);
            for(int i=0;i<left.size();i++)
            {
                for(int j=0;j<right.size();j++)
                {
                    TreeNode* p=new TreeNode(mid);
                    p->left=left[i];
                    p->right=right[j];
                    res.push_back(p);
                }
            }
        }
        return res;
    }
    vector<TreeNode*> generateTrees(int n) {
        vector<TreeNode*> res,next;
        res=dfs(1,n);
        return res;
    }
};

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原始发表：2016-10-07 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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