LeetCode OJ - Unique Binary Search Trees II

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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

分析：DFS解决，根据BST的性质 left < root < right，故递归时以root为中间点，可将整个树分为两个部分。下面的递归过程中，枚举了分割点。

/** * 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 *> generate(int beg, int end) {        vector<TreeNode* > ret;        if (beg > end) {            ret.push_back(NULL);            return ret;        }                for(int i = beg; i <= end; i++) {            vector<TreeNode* > leftTree = generate(beg, i - 1);            vector<TreeNode* > rightTree = generate(i + 1, end);            for(int j = 0; j < leftTree.size(); j++) {                for(int k = 0; k < rightTree.size(); k++) {                    TreeNode *node = new TreeNode(i + 1);                    ret.push_back(node);                    node->left = leftTree[j];                    node->right = rightTree[k];                              }                 }        }                return ret;    }        vector<TreeNode *> generateTrees(int n) {        return generate(0, n - 1);    }};

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