# 【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 should look like:

```   1
\
2
\
3
\
4
\
5
\
6```

click to show hints.

Anwser 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);
}
};```

Anwser 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;
}
};```

Anwser 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;
}
}
};```

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