Question

Here is an iterative version of preorder traversal.The idea is to
push the root into a stack.
If left child is not NULL.Push the left child and root=root->left
Else root=Top of stack;Pop;push its right child and root=root->right

Its working on some trees but for some input trees it prints a duplicate value.The behavior of stack is surprising.How does it change from (4 5 6 8) to (4 6 8)?

    void preorder(node* root)
    {stack<node*> st;
     st.push(root);
     cout<<root->val<<" \n ";
     while(!st.empty())
      {if(root->left!=NULL)
         {st.push(root->left);
          cout<<st.top()->val<<" ";
         // printstack(st);
          root=root->left;
          }
       else{root=st.top();
            st.pop();
           if(root->right!=NULL)
            {st.push(root->right);
             cout<<st.top()->val<<" ";   
             //printstack(st);
              root=root->right;
             }
           }
        }
      }

Helper function printstack used to debug

     void printstack(stack<node*> s)
      {stack<node*> t;
      t=s;
      while(!t.empty())
       {cout<<t.top()->val<<" ";
       t.pop();
       }
      } 

For this input tree,the printed stack is below.
The preorder output is
8 3 1 6 5 4 4 7 9

                  ______8_
                 /                \
             __3_          9
            /        \
            1      ___6
                     /    \
                     5      7
                    /
                   4

3 8

1 3 8

6 8

5 6 8

4 5 6 8

4 6 8 //How is the Top of stack 4??It should have been 5,Stack being ( 5 6 8)

7 8

  9

Answer 1

The issue crops up on nodes that have left children but not right. When the node has a right child, that last if block changes the root before the next iteration, but without it, the root doesn’t switch. Then, the next time through, we see a root (which has already been popped of the stack) that has a left child, so it’s left child is pushed on the stack and visited again.

Hope that helps.