Increasing Order Search Tree Solutions in C++

/// Source : https://leetcode.com/problems/increasing-order-search-tree/description/
/// Author : liuyubobobo
/// Time   : 2018-09-01

#include <iostream>
#include <vector>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


/// Inorder traversal and store all the nodes
/// Time Complexity: O(n)
/// Space Complexity: O(n)
class Solution {
public:
    TreeNode* increasingBST(TreeNode* root) {

        if(root == NULL)
            return NULL;

        vector<TreeNode*> nodes;
        inOrder(root, nodes);

        for(int i = 1; i < nodes.size(); i ++){
            nodes[i - 1]->left = NULL;
            nodes[i - 1]->right = nodes[i];
        }
        nodes.back()->left = nodes.back()->right = NULL;
        return nodes[0];
    }

private:
    void inOrder(TreeNode* node, vector<TreeNode*>& nodes){

        if(node == NULL)
            return;

        inOrder(node->left, nodes);
        nodes.push_back(node);
        inOrder(node->right, nodes);
    }
};


int main() {

    return 0;
}

/// Source : https://leetcode.com/problems/increasing-order-search-tree/description/
/// Author : liuyubobobo
/// Time   : 2018-09-01

#include <iostream>
#include <vector>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


/// Inorder traversal and link during the traversal
/// Time Complexity: O(n)
/// Space Complexity: O(h) where h is the height of the tree
class Solution {
public:
    TreeNode* increasingBST(TreeNode* root) {

        if(root == NULL)
            return NULL;

        TreeNode* tail;
        return inOrder(root, tail);
    }

private:
    TreeNode* inOrder(TreeNode* node, TreeNode* &tail){

        if(node->left == NULL && node->right == NULL){
            tail = node;
            return node;
        }

        TreeNode* ret;
        if(node->left){
            ret = inOrder(node->left, tail);
            tail->left = NULL;
            tail->right = node;
        }
        else
            ret = node;

        node->left = NULL;
        tail = node;
        if(node->right) {
            TreeNode *root = inOrder(node->right, tail);
            node->right = root;
        }

        return ret;
    }
};


int main() {

    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    Solution().increasingBST(root);

    return 0;
}

/// Source : https://leetcode.com/problems/increasing-order-search-tree/description/
/// Author : liuyubobobo
/// Time   : 2018-09-01

#include <iostream>
#include <vector>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


/// Inorder traversal and link during the traversal
/// Using a class member TreeNode cur:)
///
/// Time Complexity: O(n)
/// Space Complexity: O(h) where h is the height of the tree
class Solution {

private:
    TreeNode* cur;

public:
    TreeNode* increasingBST(TreeNode* root) {

        if(root == NULL)
            return NULL;

        TreeNode* dummyRoot = new TreeNode(-1);
        cur = dummyRoot;
        inOrder(root);

        TreeNode* ret = dummyRoot->right;
        delete dummyRoot;
        return ret;
    }

private:
    void inOrder(TreeNode* node){

        if(node == NULL)
            return;

        inOrder(node->left);

        cur->right = node;
        cur = cur->right;
        cur->left = NULL;

        inOrder(node->right);
    }
};


int main() {

    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    Solution().increasingBST(root);

    return 0;
}

/// Source : https://leetcode.com/problems/increasing-order-search-tree/description/
/// Author : liuyubobobo
/// Time   : 2018-09-02

#include <iostream>
#include <vector>

using namespace std;

/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


/// Using Morris Treversal
/// Time Complexity: O(n)
/// Space Complexity: O(1)
class Solution {

private:
    TreeNode* cur;

public:
    TreeNode* increasingBST(TreeNode* root) {

        if(root == NULL)
            return NULL;

        TreeNode* dummyRoot = new TreeNode(-1);
        cur = dummyRoot;

        // Morris Traversal
        TreeNode* node = root;
        while(node){
            if(node->left == NULL){
                cur->right = node;
                cur = cur->right;
                cur->left = NULL;

                node = node->right;
            }
            else{
                TreeNode* prev = node->left;
                while(prev->right && prev->right != node)
                    prev = prev->right;

                if(prev->right == NULL){
                    prev->right = node;
                    node = node->left;
                }
                else{
                    prev->right = NULL;

                    cur->right = node;
                    cur = cur->right;
                    cur->left = NULL;

                    node = node->right;
                }
            }
        }

        TreeNode* ret = dummyRoot->right;
        delete dummyRoot;
        return ret;
    }
};


int main() {

    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    Solution().increasingBST(root);

    return 0;
}