[Leetcode] flatten binary tree to linked list

Question:

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

         1        / \       2   5      / \   \     3   4   6

The flattened tree shoshould look like:

   1    \     2      \       3        \         4          \           5            \             6

Click to show hints.

An1_1:

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    void flatten(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (root == NULL) return;        TreeNode* left = root->left;        TreeNode* right = root->right;      // the original right child        if (left) {            root->right = left;            root->left = NULL;            TreeNode* rightmost = left;            while(rightmost->right) {                rightmost = rightmost->right;                            }            rightmost->right = right; // point to the original right child        }        flatten(root->right);        }};

An1_2:

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    void flatten(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (root == NULL) return;                if (root->left)            flatten(root->left);                    if (root->right)            flatten(root->right);                    if (NULL == root->left)            return;                    TreeNode ** ptn = &(root->left->right);        while (*ptn) {            ptn = & ((*ptn)->right);        }                    *ptn = root->right;        root->right = root->left;   // link right to left        root->left = NULL;    }};

An1_3: 

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    void flatten(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        bool flag = true;        stack<TreeNode *> t;        TreeNode *pre = NULL;                if(root) {            t.push(root);        }                while(t.size()) {            TreeNode *cur = t.top();            if(flag) {                if(cur->left && cur->left != pre) {                    t.push(cur->left);                } else {                    flag = false;                }            } else {                if(cur->right && cur->right != pre) {                    t.push(cur->right);                    flag = true;                } else {                    t.pop();                    //at this time, the sub-tree of cur is flattened, so just flatten the cur                    TreeNode *left = cur->left;                    if(left) {                        TreeNode *lastLeft = left;                        while(lastLeft->right) {                            lastLeft = lastLeft->right;                        }                        lastLeft->right = cur->right;                        cur->right = left;                        cur->left = NULL;                    }                }            }            pre = cur;        }    }};

Reference recommendations: 

Leetcode problem: flatten binary tree to linked list