Flatten Binary Tree to Linked List (Stack, tree; DFS)

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 method I: Stack

structTreeNode {intVal; TreeNode*Left ; TreeNode*Right ; TreeNode (intx): Val (x), left (null), right (null) {}};classSolution { Public:    voidFlatten (TreeNode *root) {        if(!root)return; TreeNode* Flatroot =NewTreeNode (root->val); TreeNode* Flatnode =Flatroot; TreeNode*Current ; Stack<TreeNode*>Tree_stack; //stack is first into the back, so the reverse order into the stack, that is, first right son, then left son        if(root->Right ) {Tree_stack.push (Root-Right ); }                    if(root->Left ) {Tree_stack.push (Root-Left ); }         while(!Tree_stack.empty ()) { Current=Tree_stack.top ();                        Tree_stack.pop (); Flatnode->right =NewTreeNode (current->val); Flatnode= flatnode->Right ; //Treat the stack element as the root node and put its right son and left son on the stack again            if(current->Right ) {Tree_stack.push ( current-Right ); }                        if(current->Left ) {Tree_stack.push ( current-Left ); }              }                *root = *Flatroot; }};

Method II: Recursion, pre-sequence traversal

classSolution { Public:    voidFlatten (TreeNode *root) {        if(!root)return; TreeNode* Newroot =NewTreeNode (root->val);        Preordertraverse (Root, Newroot); *root = *Newroot; } TreeNode* Preordertraverse (TreeNode *root, TreeNode *newroot) {        if(root->Left ) {Newroot->right =NewTreeNode (root->left->val); Newroot= Preordertraverse (Root->left, newroot->Right ); }        if(root->Right ) {Newroot->right =NewTreeNode (root->right->val); Newroot= Preordertraverse (Root->right, newroot->Right ); }        returnNewroot; }};

Flatten Binary Tree to Linked List (Stack, tree; DFS)