145. Binary Tree Postorder Traversal (Stack, tree)

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

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

   1         2    /   3

Return [3,2,1] .

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

classSolution { Public: Vector<int> Postordertraversal (TreeNode *root) {Vector<int>result; if(!root)returnresult; Stack<MyNode*>Treestack; Mynode* Myroot =NewMynode (root); Mynode* Current, *NewNode;              Treestack.push (Myroot);  while(!Treestack.empty ()) { Current=Treestack.top ();            Treestack.pop (); if(!current->flag) { Current->flag =true;                Treestack.push (current); if(current->node->Right ) {NewNode=NewMynode (current->node->Right );                Treestack.push (NewNode); }                if(current->node->Left ) {NewNode=NewMynode (current->node->Left );                Treestack.push (NewNode); }            }            Else{result.push_back ( current->node->val); }        }        returnresult; }    structMynode {TreeNode*node; BOOLFlag//indicate if the node has been visitedMynode (treenode* x): Node (x), Flag (false) {}    };};

145. Binary Tree Postorder Traversal (Stack, tree)