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Iterative Boundary Traversal of Complete Binary tree
  • Difficulty Level : Hard
  • Last Updated : 28 Dec, 2020

Given a complete binary tree, traverse it such that all the boundary nodes are visited in Anti-Clockwise order starting from the root.
Example: 
 

Input:
               18
           /       \  
         15         30  
        /  \        /  \
      40    50    100   20
Output: 18 15 40 50 100 20 30

 

Approach: 
 

  • Traverse left-most nodes of the tree from top to down. (Left boundary)
  • Traverse bottom-most level of the tree from left to right. (Leaf nodes)
  • Traverse right-most nodes of the tree from bottom to up. (Right boundary)

We can traverse the left boundary quite easily with the help of a while loop that checks when the node doesn’t have any left child. Similarly, we can traverse the right boundary quite easily with the help of a while loop that checks when the node doesn’t have any right child.
The main challenge here is to traverse the last level of the tree in left to right order. To traverse level-wise there is BFS and order of left to right can be taken care of by pushing left nodes in the queue first. So the only thing left now is to make sure it is the last level. Just check whether the node has any child and only include them. 
We will have to take special care of the corner case that same nodes are not traversed again. In the example above 40 is a part of the left boundary as well as leaf nodes. Similarly, 20 is a part of the right boundary as well as leaf nodes. 
So we will have to traverse only till the second last node of both the boundaries in that case. Also keep in mind we should not traverse the root again.
Below is the implementation of the above approach: 
 

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree node has data, pointer
   to left child and a pointer to right
   child */
struct Node {
    int data;
    struct Node *left, *right;
};
 
/* Helper function that allocates a new
   node with the given data and NULL left
   and right pointers. */
struct Node* newNode(int data)
{
    Node* temp = new Node;
 
    temp->data = data;
    temp->left = temp->right = NULL;
 
    return temp;
}
 
// Function to print the nodes of a complete
// binary tree in boundary traversal order
void boundaryTraversal(Node* root)
{
    if (root) {
 
        // If there is only 1 node print it
        // and return
        if (!(root->left) && !(root->right)) {
            cout << root->data << endl;
            return;
        }
 
        // List to store order of traversed
        // nodes
        vector<Node*> list;
        list.push_back(root);
 
        // Traverse left boundary without root
        // and last node
        Node* L = root->left;
        while (L->left) {
            list.push_back(L);
            L = L->left;
        }
 
        // BFS designed to only include leaf nodes
        queue<Node*> q;
        q.push(root);
        while (!q.empty()) {
            Node* temp = q.front();
            q.pop();
            if (!(temp->left) && !(temp->right)) {
                list.push_back(temp);
            }
            if (temp->left) {
                q.push(temp->left);
            }
            if (temp->right) {
                q.push(temp->right);
            }
        }
 
        // Traverse right boundary without root
        // and last node
        vector<Node*> list_r;
        Node* R = root->right;
        while (R->right) {
            list_r.push_back(R);
            R = R->right;
        }
 
        // Reversing the order
        reverse(list_r.begin(), list_r.end());
 
        // Concatenating the two lists
        list.insert(list.end(), list_r.begin(),
                                 list_r.end());
 
        // Printing the node's data from the list
        for (auto i : list) {
            cout << i->data << " ";
        }
        cout << endl;
        return;
    }
}
 
// Driver code
int main()
{
 
    // Root node of the tree
    Node* root = newNode(20);
 
    root->left = newNode(8);
    root->right = newNode(22);
 
    root->left->left = newNode(4);
    root->left->right = newNode(12);
 
    root->right->left = newNode(10);
    root->right->right = newNode(25);
 
    boundaryTraversal(root);
 
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
 
class GFG{
 
/* A binary tree node has data, pointer
   to left child and a pointer to right
   child */
static class Node {
    int data;
    Node left, right;
};
 
/* Helper function that allocates a new
   node with the given data and null left
   and right pointers. */
static Node newNode(int data)
{
    Node temp = new Node();
 
    temp.data = data;
    temp.left = temp.right = null;
 
    return temp;
}
 
// Function to print the nodes of a complete
// binary tree in boundary traversal order
static void boundaryTraversal(Node root)
{
    if (root != null) {
 
        // If there is only 1 node print it
        // and return
        if ((root.left == null) && (root.right == null)) {
            System.out.print(root.data +"\n");
            return;
        }
 
        // List to store order of traversed
        // nodes
        Vector<Node> list = new Vector<Node>();
        list.add(root);
 
        // Traverse left boundary without root
        // and last node
        Node L = root.left;
        while (L.left != null) {
            list.add(L);
            L = L.left;
        }
 
        // BFS designed to only include leaf nodes
        Queue<Node> q = new LinkedList<>();
        q.add(root);
        while (!q.isEmpty()) {
            Node temp = q.peek();
            q.remove();
            if ((temp.left == null) && (temp.right == null)) {
                list.add(temp);
            }
            if (temp.left != null) {
                q.add(temp.left);
            }
            if (temp.right != null) {
                q.add(temp.right);
            }
        }
 
        // Traverse right boundary without root
        // and last node
        Vector<Node> list_r = new Vector<Node>();
        Node R = root.right;
        while (R.right != null) {
            list_r.add(R);
            R = R.right;
        }
 
        // Reversing the order
        Collections.reverse(list_r);
 
        // Concatenating the two lists
        list.addAll(list_r);
 
        // Printing the node's data from the list
        for (Node i : list) {
            System.out.print(i.data + " ");
        }
        System.out.println();
        return;
    }
}
 
// Driver code
public static void main(String[] args)
{
 
    // Root node of the tree
    Node root = newNode(20);
 
    root.left = newNode(8);
    root.right = newNode(22);
 
    root.left.left = newNode(4);
    root.left.right = newNode(12);
 
    root.right.left = newNode(10);
    root.right.right = newNode(25);
 
    boundaryTraversal(root);
}
}
 
// This code is contributed by Princi Singh

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Python

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# Python implementation of the approach
from collections import deque
  
# A binary tree node
class Node:
  
    # A constructor for making a new node
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
  
# Function to print the nodes of a complete
# binary tree in boundary traversal order
def boundaryTraversal(root):
    # If there is only 1 node print it and return
    if root:
        if not root.left and not root.right:
            print (root.data)
            return
  
        # List to store order of traversed nodes
        list = []
        list.append(root)
  
        # Traverse left boundary without root
        # and last node
        temp = root.left
        while temp.left:
            list.append(temp)
            temp = temp.left
  
        # BFS designed to only include leaf nodes
        q = deque()
        q.append(root)
        while len(q) != 0:
            x = q.pop()
            if not x.left and not x.right:
                list.append(x)
            if x.right:
                q.append(x.right)
            if x.left:
                q.append(x.left)
  
        # Traverse right boundary without root
        # and last node
        list_r = []
        temp = root.right
        while temp.right:
            list.append(temp)
            temp = temp.right
  
        # Reversing the order
        list_r = list_r[::-1]
  
        # Concatenating the two lists
        list += list_r
  
        # Printing the node's data from the list
        print (" ".join([str(i.data) for i in list]))
    return
  
# Root node of the tree
  
root = Node(20)
  
root.left = Node(8)
root.right = Node(22)
  
root.left.left = Node(4)
root.left.right = Node(12)
 
root.right.left = Node(10)
root.right.right = Node(25)
  
boundaryTraversal(root)

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
 
/* A binary tree node has data, pointer
   to left child and a pointer to right
   child */
class Node
{
    public int data;
    public Node left, right;
};
 
/* Helper function that allocates a new
   node with the given data and null left
   and right pointers. */
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
 
// Function to print the nodes of a complete
// binary tree in boundary traversal order
static void boundaryTraversal(Node root)
{
    if (root != null)
    {
 
        // If there is only 1 node print it
        // and return
        if ((root.left == null) &&
            (root.right == null))
        {
            Console.Write(root.data + "\n");
            return;
        }
 
        // List to store order of traversed
        // nodes
        List<Node> list = new List<Node>();
        list.Add(root);
 
        // Traverse left boundary without root
        // and last node
        Node L = root.left;
        while (L.left != null)
        {
            list.Add(L);
            L = L.left;
        }
 
        // BFS designed to only include leaf nodes
        Queue<Node> q = new Queue<Node>();
        q.Enqueue(root);
        while (q.Count != 0)
        {
            Node temp = q.Peek();
            q.Dequeue();
            if ((temp.left == null) &&
                (temp.right == null))
            {
                list.Add(temp);
            }
            if (temp.left != null)
            {
                q.Enqueue(temp.left);
            }
            if (temp.right != null)
            {
                q.Enqueue(temp.right);
            }
        }
 
        // Traverse right boundary without root
        // and last node
        List<Node> list_r = new List<Node>();
        Node R = root.right;
        while (R.right != null)
        {
            list_r.Add(R);
            R = R.right;
        }
 
        // Reversing the order
        list_r.Reverse();
 
        // Concatenating the two lists
        list.InsertRange(list.Count-1, list_r);
 
        // Printing the node's data from the list
        foreach (Node i in list)
        {
            Console.Write(i.data + " ");
        }
        Console.WriteLine();
        return;
    }
}
 
// Driver code
public static void Main(String[] args)
{
 
    // Root node of the tree
    Node root = newNode(20);
    root.left = newNode(8);
    root.right = newNode(22);
    root.left.left = newNode(4);
    root.left.right = newNode(12);
    root.right.left = newNode(10);
    root.right.right = newNode(25);
    boundaryTraversal(root);
}
}
 
// This code is contributed by 29AjayKumar

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Output: 

20 8 4 12 10 25 22

 

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