Leetcode 144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes ' values.

For example:
Given binary Tree {1,#,2,3} ,

   1         2    /   3

Return [1,2,3] .

Note:recursive solution is trivial, could do it iteratively?

The recursive approach is simple:

1vector<int> Preordertraversal (TreeNode *root)2     {3vector<int>Vtree;4 preorder (vtree, root);5         returnVtree;6     }7     8     voidPreorder (vector<int> &vtree, TreeNode *root)9     {Ten         if(Root = =NULL) One             return; AVtree.push_back (root->val); -Preorder (Vtree, root->Left ); -Preorder (Vtree, root->Right ); the}

For non-recursive methods, a stack is needed to temporarily store the tree node

1) If current node p isn't empty, visit it and put P->right in stack, then P = P->left

2) Else, get the top node in stack

1vector<int> Preordertraversal (TreeNode *root)2     {3vector<int>Vtree;4Stack<treenode *>Stree;5TreeNode *visit =Root;6         7          while(!stree.empty () | | (Visit! =NULL))8         {9             if(Visit! =NULL)Ten             {   OneVtree.push_back (Visit->val);//If visit is not empty, visit it A                 if(Visit->right! = NULL)//if Visit->right exit, push it in stack -Stree.push (visit->Right ); -Visit = visit->Left ; the             } -             Else    //visit is null -             { -Visit =stree.top (); + Stree.pop (); -             } +         } A          at         returnVtree; -}

Leetcode 144. Binary Tree Preorder Traversal