Minimum Distance Between BST Nodes Solutions in C++

/// Source : https://leetcode.com/problems/minimum-distance-between-bst-nodes/description/
/// Author : liuyubobobo
/// Time   : 2010-02-10

#include <iostream>
#include <cassert>

using namespace std;

/// Brute Force
/// For every node, find his prev and next node and compare the diff
///
/// Time Complexity: O(n^2)
/// Space Complexity: O(logn)

/// 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:
    int minDiffInBST(TreeNode* root) {

        if(root == NULL)
            return INT_MAX;

        if(root->left == NULL && root->right == NULL)
            return INT_MAX;

        int diff1 = root->left ? root->val - maxInBST(root->left) : INT_MAX;
        int diff2 = root->right ? minInBST(root->right) - root->val : INT_MAX;

        return min(min(diff1, diff2),
                   min(minDiffInBST(root->left), minDiffInBST(root->right)));
    }

private:
    int maxInBST(TreeNode* node){
        assert(node != NULL);
        if(node->right)
            return maxInBST(node->right);
        return node->val;
    }

    int minInBST(TreeNode* node){
        assert(node != NULL);
        if(node->left)
            return minInBST(node->left);
        return node->val;
    }
};

int main() {

    // [90,69,null,49,89,null,52,null,null,null,null]

    return 0;
}

/// Source : https://leetcode.com/problems/minimum-distance-between-bst-nodes/description/
/// Author : liuyubobobo
/// Time   : 2010-02-11

#include <iostream>
#include <vector>
#include <algorithm>
#include <cassert>

using namespace std;

/// Using Array
/// Store All Numbers in an array and sort, then get the min distance
///
/// Time Complexity: O(nlogn)
/// Space Complexity: O(n)

/// 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:
    int minDiffInBST(TreeNode* root) {

        vector<int> nums;
        dfs(root, nums);
        sort(nums.begin(), nums.end());

        int res = INT_MAX;
        for(int i = 1 ; i < nums.size() ; i ++)
            res = min(res, nums[i] - nums[i-1]);

        return res;
    }

private:
    void dfs(TreeNode* node, vector<int>& nums){

        if(node == NULL)
            return;

        nums.push_back(node->val);
        dfs(node->left, nums);
        dfs(node->right, nums);
    }
};

int main() {

    // [90,69,null,49,89,null,52,null,null,null,null]

    return 0;
}

/// Source : https://leetcode.com/problems/minimum-distance-between-bst-nodes/description/
/// Author : liuyubobobo
/// Time   : 2010-02-11

#include <iostream>
#include <vector>
#include <algorithm>
#include <cassert>

using namespace std;

/// Using Array
/// Inorder traverse the tree and put every number in an array.
/// No need to sort the array.
///
/// Time Complexity: O(n)
/// Space Complexity: O(n)

/// 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:
    int minDiffInBST(TreeNode* root) {

        vector<int> nums;
        dfs(root, nums);

        int res = INT_MAX;
        for(int i = 1 ; i < nums.size() ; i ++)
            res = min(res, nums[i] - nums[i-1]);

        return res;
    }

private:
    void dfs(TreeNode* node, vector<int>& nums){

        if(node == NULL)
            return;

        dfs(node->left, nums);
        nums.push_back(node->val);
        dfs(node->right, nums);
    }
};

int main() {

    // [90,69,null,49,89,null,52,null,null,null,null]

    return 0;
}

/// Source : https://leetcode.com/problems/minimum-distance-between-bst-nodes/description/
/// Author : liuyubobobo
/// Time   : 2010-02-11

#include <iostream>
#include <vector>
#include <algorithm>
#include <cassert>

using namespace std;

/// Inorder Traverse
/// Inorder traverse the tree and record every neighbor numbers diff during the process.
///
/// Time Complexity: O(n)
/// Space Complexity: O(logn)

/// 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 {

private:
    int* lastNumber;

public:
    int minDiffInBST(TreeNode* root) {

        int res = INT_MAX;
        lastNumber = NULL;
        dfs(root, res);

        return res;
    }

private:
    void dfs(TreeNode* node, int& res){

        if(node == NULL)
            return;

        dfs(node->left, res);

        if(lastNumber != NULL){
            // cout << node->val << " " << (*lastNumber) << endl;
            res = min(res, node->val - *lastNumber);
        }

        lastNumber = &(node->val);

        dfs(node->right, res);
    }
};

int main() {

    // [90,69,null,49,89,null,52,null,null,null,null]

    // [27,null,34,null,58,50,null,44,null,null,null]
    TreeNode* node44 = new TreeNode(44);
    TreeNode* node50 = new TreeNode(50);
    TreeNode* node58 = new TreeNode(58);
    TreeNode* node34 = new TreeNode(34);
    TreeNode* node27 = new TreeNode(27);

    node27->right = node34;
    node34->right = node58;
    node58->left = node50;
    node50->left = node44;

    cout << Solution().minDiffInBST(node27);

    return 0;
}