To print the boundary of Binary Tree

I guess this should do the trick:

traverse(BinaryTree *root)
{
  if (!root) return;
  cout << p->data << " ";
  if (root->left ) traverseL(root->left ); //special function for outer left
  if (root->right) traverseR(root->right); //special function for outer right
}

traverseL(BinaryTree *p)
{
  cout << p->data << " ";
  if (root->left ) traverseL(root->left ); //still in outer left
  if (root->right) traverseC(root->right); 
}

traverseR(BinaryTree *p)
{
  if (root->left ) traverseC(root->left );
  if (root->right) traverseR(root->right); //still in outer right
  cout << p->data << " ";
}

traverseC(BinaryTree *p)
{
  if (!root->left && !root->right) //bottom reached
    cout << p->data << " ";
  else
  {
    if (root->left ) traverseC(root->left );
    if (root->right) traverseC(root->right);
  }
}

Start with the traverse function. Got rid of the null-queries at the beginning of each method (avoids one function call at each end). Does not need bool variables, simply uses three different traversal methods:

• one for the left edge, outputting the node before traversal
• one for the right edge, outputting the node after traversal
• one for all other nodes, outputting the node if there are no siblings.

Below is a recursive solution in Python3 with time complexity O(n). Algorithm here is to print left most nodes from top to bottom, leaf nodes from left to right and right most nodes from bottom to top. Adding boolean flags (isLeft,isRight) for left and right tree traversal simplifies the code and drives the time complexity of O(n).

#Print tree boundary nodes
def TreeBoundry(node,isLeft,isRight):
    #Left most node and leaf nodes
    if(isLeft or isLeaf(node)): print(node.data,end=' ')
    #Process next left node
    if(node.getLeft() is not None): TreeBoundry(node.getLeft(), True,False)
    #Process next right node
    if(node.getRight() is not None):TreeBoundry(node.getRight(), False,True)
    #Right most node
    if(isRight and not isLeft and  not isLeaf(node)):print(node.data,end=' ')

#Check is a node is leaf
def isLeaf(node):
   if (node.getLeft() is None and  node.getRight() is None):
       return True
   else:
       return False

#Get started
#https://github.com/harishvc/challenges/blob/master/binary-tree-edge-nodes.py
TreeBoundry(root,True,True)