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Boundary Traversal of binary tree

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  • Difficulty Level : Medium
  • Last Updated : 11 Jan, 2023
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Given a binary tree, print boundary nodes of the binary tree Anti-Clockwise starting from the root.

 The boundary includes 

  1. left boundary (nodes on left excluding leaf nodes)
  2. leaves (consist of only the leaf nodes)
  3. right boundary (nodes on right excluding leaf nodes)

The left boundary is defined as the path from the root to the left-most node. The right boundary is defined as the path from the root to the right-most node. If the root doesn’t have left subtree or right subtree, then the root itself is left boundary or right boundary. Note this definition only applies to the input binary tree, and not apply to any subtrees.
The left-most node is defined as a leaf node you could reach when you always firstly travel to the left subtree if it exists. If not, travel to the right subtree. Repeat until you reach a leaf node.
The right-most node is also defined in the same way with left and right exchanged. 
For example, boundary traversal of the following tree is “20 8 4 10 14 25 22”

This is how we write the traversal:

root : 20

left- boundary nodes: 8

leaf nodes: 4 10 14 25

right – boundary nodes: 22 

 


 

We break the problem in 3 parts: 

1. Print the left boundary in top-down manner. 
2. Print all leaf nodes from left to right, which can again be sub-divided into two sub-parts: 
…..2.1 Print all leaf nodes of left sub-tree from left to right. 
…..2.2 Print all leaf nodes of right subtree from left to right. 
3. Print the right boundary in bottom-up manner.
We need to take care of one thing that nodes are not printed again. e.g. The left most node is also the leaf node of the tree.
Based on the above cases, below is the implementation:
 

Implementation:

C++




#include <iostream>
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;
};
 
// Utility function to create a new tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = nullptr;
    return temp;
}
 
// A simple function to print leaf nodes of a binary tree
void printLeaves(Node* root)
{
    if (root == nullptr)
        return;
 
    printLeaves(root->left);
 
    // Print it if it is a leaf node
    if (!(root->left) && !(root->right))
        cout << root->data << " ";
 
    printLeaves(root->right);
}
 
// A function to print all left boundary nodes, except a
// leaf node. Print the nodes in TOP DOWN manner
void printBoundaryLeft(Node* root)
{
    if (root == nullptr)
        return;
 
    if (root->left) {
 
        // to ensure top down order, print the node
        // before calling itself for left subtree
        cout << root->data << " ";
        printBoundaryLeft(root->left);
    }
    else if (root->right) {
        cout << root->data << " ";
        printBoundaryLeft(root->right);
    }
    // do nothing if it is a leaf node, this way we avoid
    // duplicates in output
}
 
// A function to print all right boundary nodes, except a
// leaf node Print the nodes in BOTTOM UP manner
void printBoundaryRight(Node* root)
{
    if (root == nullptr)
        return;
 
    if (root->right) {
        // to ensure bottom up order, first call for right
        // subtree, then print this node
        printBoundaryRight(root->right);
        cout << root->data << " ";
    }
    else if (root->left) {
        printBoundaryRight(root->left);
        cout << root->data << " ";
    }
    // do nothing if it is a leaf node, this way we avoid
    // duplicates in output
}
 
// A function to do boundary traversal of a given binary
// tree
void printBoundary(Node* root)
{
    if (root == nullptr)
        return;
 
    cout << root->data << " ";
 
    // Print the left boundary in top-down manner.
    printBoundaryLeft(root->left);
 
    // Print all leaf nodes
    printLeaves(root->left);
    printLeaves(root->right);
 
    // Print the right boundary in bottom-up manner
    printBoundaryRight(root->right);
}
 
// Driver program to test above functions
int main()
{
    // Let us construct the tree given in the above diagram
    Node* root = newNode(20);
    root->left = newNode(8);
    root->left->left = newNode(4);
    root->left->right = newNode(12);
    root->left->right->left = newNode(10);
    root->left->right->right = newNode(14);
    root->right = newNode(22);
    root->right->right = newNode(25);
 
    printBoundary(root);
 
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

C




/* C program for boundary traversal
of a binary tree */
 
#include <stdio.h>
#include <stdlib.h>
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node {
    int data;
    struct node *left, *right;
};
 
// A simple function to print leaf nodes of a binary tree
void printLeaves(struct node* root)
{
    if (root == NULL)
        return;
 
    printLeaves(root->left);
 
    // Print it if it is a leaf node
    if (!(root->left) && !(root->right))
        printf("%d ", root->data);
 
    printLeaves(root->right);
}
 
// A function to print all left boundary nodes, except a leaf node.
// Print the nodes in TOP DOWN manner
void printBoundaryLeft(struct node* root)
{
    if (root == NULL)
        return;
 
    if (root->left) {
 
        // to ensure top down order, print the node
        // before calling itself for left subtree
        printf("%d ", root->data);
        printBoundaryLeft(root->left);
    }
    else if (root->right) {
        printf("%d ", root->data);
        printBoundaryLeft(root->right);
    }
    // do nothing if it is a leaf node, this way we avoid
    // duplicates in output
}
 
// A function to print all right boundary nodes, except a leaf node
// Print the nodes in BOTTOM UP manner
void printBoundaryRight(struct node* root)
{
    if (root == NULL)
        return;
 
    if (root->right) {
        // to ensure bottom up order, first call for right
        // subtree, then print this node
        printBoundaryRight(root->right);
        printf("%d ", root->data);
    }
    else if (root->left) {
        printBoundaryRight(root->left);
        printf("%d ", root->data);
    }
    // do nothing if it is a leaf node, this way we avoid
    // duplicates in output
}
 
// A function to do boundary traversal of a given binary tree
void printBoundary(struct node* root)
{
    if (root == NULL)
        return;
 
    printf("%d ", root->data);
 
    // Print the left boundary in top-down manner.
    printBoundaryLeft(root->left);
 
    // Print all leaf nodes
    printLeaves(root->left);
    printLeaves(root->right);
 
    // Print the right boundary in bottom-up manner
    printBoundaryRight(root->right);
}
 
// A utility function to create a node
struct node* newNode(int data)
{
    struct node* temp = (struct node*)malloc(sizeof(struct node));
 
    temp->data = data;
    temp->left = temp->right = NULL;
 
    return temp;
}
 
// Driver program to test above functions
int main()
{
    // Let us construct the tree given in the above diagram
    struct node* root = newNode(20);
    root->left = newNode(8);
    root->left->left = newNode(4);
    root->left->right = newNode(12);
    root->left->right->left = newNode(10);
    root->left->right->right = newNode(14);
    root->right = newNode(22);
    root->right->right = newNode(25);
 
    printBoundary(root);
    return 0;
}
// This code has been contributed by Aditya Kumar (adityakumar129)

Java




// Java program to print boundary traversal of binary tree
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
class Node {
    int data;
    Node left, right;
 
    Node(int item)
    {
        data = item;
        left = right = null;
    }
}
 
class BinaryTree {
    Node root;
 
    // A simple function to print leaf nodes of a binary tree
    void printLeaves(Node node)
    {
        if (node == null)
            return;
 
        printLeaves(node.left);
        // Print it if it is a leaf node
        if (node.left == null && node.right == null)
            System.out.print(node.data + " ");
        printLeaves(node.right);
    }
 
    // A function to print all left boundary nodes, except a leaf node.
    // Print the nodes in TOP DOWN manner
    void printBoundaryLeft(Node node)
    {
        if (node == null)
            return;
 
        if (node.left != null) {
            // to ensure top down order, print the node
            // before calling itself for left subtree
            System.out.print(node.data + " ");
            printBoundaryLeft(node.left);
        }
        else if (node.right != null) {
            System.out.print(node.data + " ");
            printBoundaryLeft(node.right);
        }
 
        // do nothing if it is a leaf node, this way we avoid
        // duplicates in output
    }
 
    // A function to print all right boundary nodes, except a leaf node
    // Print the nodes in BOTTOM UP manner
    void printBoundaryRight(Node node)
    {
        if (node == null)
            return;
 
        if (node.right != null) {
            // to ensure bottom up order, first call for right
            // subtree, then print this node
            printBoundaryRight(node.right);
            System.out.print(node.data + " ");
        }
        else if (node.left != null) {
            printBoundaryRight(node.left);
            System.out.print(node.data + " ");
        }
        // do nothing if it is a leaf node, this way we avoid
        // duplicates in output
    }
 
    // A function to do boundary traversal of a given binary tree
    void printBoundary(Node node)
    {
        if (node == null)
            return;
 
        System.out.print(node.data + " ");
 
        // Print the left boundary in top-down manner.
        printBoundaryLeft(node.left);
 
        // Print all leaf nodes
        printLeaves(node.left);
        printLeaves(node.right);
 
        // Print the right boundary in bottom-up manner
        printBoundaryRight(node.right);
    }
 
    // Driver program to test above functions
    public static void main(String args[])
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(20);
        tree.root.left = new Node(8);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(12);
        tree.root.left.right.left = new Node(10);
        tree.root.left.right.right = new Node(14);
        tree.root.right = new Node(22);
        tree.root.right.right = new Node(25);
        tree.printBoundary(tree.root);
    }
}

Python3




# Python3 program for binary traversal of binary tree
 
# A binary tree node
class Node:
 
    # Constructor to create a new node
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# A simple function to print leaf nodes of a Binary Tree
def printLeaves(root):
    if(root):
        printLeaves(root.left)
         
        # Print it if it is a leaf node
        if root.left is None and root.right is None:
            print(root.data),
 
        printLeaves(root.right)
 
# A function to print all left boundary nodes, except a
# leaf node. Print the nodes in TOP DOWN manner
def printBoundaryLeft(root):
     
    if(root):
        if (root.left):
             
            # to ensure top down order, print the node
            # before calling itself for left subtree
            print(root.data)
            printBoundaryLeft(root.left)
         
        elif(root.right):
            print (root.data)
            printBoundaryLeft(root.right)
         
        # do nothing if it is a leaf node, this way we
        # avoid duplicates in output
 
 
# A function to print all right boundary nodes, except
# a leaf node. Print the nodes in BOTTOM UP manner
def printBoundaryRight(root):
     
    if(root):
        if (root.right):
            # to ensure bottom up order, first call for
            # right subtree, then print this node
            printBoundaryRight(root.right)
            print(root.data)
         
        elif(root.left):
            printBoundaryRight(root.left)
            print(root.data)
 
        # do nothing if it is a leaf node, this way we
        # avoid duplicates in output
 
 
# A function to do boundary traversal of a given binary tree
def printBoundary(root):
    if (root):
        print(root.data)
         
        # Print the left boundary in top-down manner
        printBoundaryLeft(root.left)
 
        # Print all leaf nodes
        printLeaves(root.left)
        printLeaves(root.right)
 
        # Print the right boundary in bottom-up manner
        printBoundaryRight(root.right)
 
 
# Driver program to test above function
root = Node(20)
root.left = Node(8)
root.left.left = Node(4)
root.left.right = Node(12)
root.left.right.left = Node(10)
root.left.right.right = Node(14)
root.right = Node(22)
root.right.right = Node(25)
printBoundary(root)
 
# This code is contributed by
# Nikhil Kumar Singh(nickzuck_007)

C#




// C# program to print boundary traversal
// of binary tree
using System;
 
/* A binary tree node has data,
pointer to left child and a pointer
to right child */
public class Node {
    public int data;
    public Node left, right;
 
    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}
 
class GFG {
    public Node root;
 
    // A simple function to print leaf
    // nodes of a binary tree
    public virtual void printLeaves(Node node)
    {
        if (node == null)
            return;
 
        printLeaves(node.left);
 
        // Print it if it is a leaf node
        if (node.left == null && node.right == null) {
            Console.Write(node.data + " ");
        }
        printLeaves(node.right);
    }
 
    // A function to print all left boundary
    // nodes, except a leaf node. Print the
    // nodes in TOP DOWN manner
    public virtual void printBoundaryLeft(Node node)
    {
        if (node == null)
            return;
 
        if (node.left != null) {
 
            // to ensure top down order, print the node
            // before calling itself for left subtree
            Console.Write(node.data + " ");
            printBoundaryLeft(node.left);
        }
        else if (node.right != null) {
            Console.Write(node.data + " ");
            printBoundaryLeft(node.right);
        }
 
        // do nothing if it is a leaf node,
        // this way we avoid duplicates in output
    }
 
    // A function to print all right boundary
    // nodes, except a leaf node. Print the
    // nodes in BOTTOM UP manner
    public virtual void printBoundaryRight(Node node)
    {
        if (node == null)
            return;
 
        if (node.right != null) {
            // to ensure bottom up order,
            // first call for right subtree,
            // then print this node
            printBoundaryRight(node.right);
            Console.Write(node.data + " ");
        }
        else if (node.left != null) {
            printBoundaryRight(node.left);
            Console.Write(node.data + " ");
        }
        // do nothing if it is a leaf node,
        // this way we avoid duplicates in output
    }
 
    // A function to do boundary traversal
    // of a given binary tree
    public virtual void printBoundary(Node node)
    {
        if (node == null)
            return;
 
        Console.Write(node.data + " ");
 
        // Print the left boundary in
        // top-down manner.
        printBoundaryLeft(node.left);
 
        // Print all leaf nodes
        printLeaves(node.left);
        printLeaves(node.right);
 
        // Print the right boundary in
        // bottom-up manner
        printBoundaryRight(node.right);
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        GFG tree = new GFG();
        tree.root = new Node(20);
        tree.root.left = new Node(8);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(12);
        tree.root.left.right.left = new Node(10);
        tree.root.left.right.right = new Node(14);
        tree.root.right = new Node(22);
        tree.root.right.right = new Node(25);
        tree.printBoundary(tree.root);
    }
}
 
// This code is contributed by Shrikant13

Javascript




<script>
     
    // JavaScript program to print boundary
    // traversal of binary tree
     
    class Node
    {
        constructor(item) {
           this.left = null;
           this.right = null;
           this.data = item;
        }
    }
     
    let root;
   
    // A simple function to print leaf nodes of a binary tree
    function printLeaves(node)
    {
        if (node == null)
            return;
   
        printLeaves(node.left);
        // Print it if it is a leaf node
        if (node.left == null && node.right == null)
            document.write(node.data + " ");
        printLeaves(node.right);
    }
   
    // A function to print all left boundary nodes,
    // except a leaf node.
    // Print the nodes in TOP DOWN manner
    function printBoundaryLeft(node)
    {
        if (node == null)
            return;
   
        if (node.left != null) {
            // to ensure top down order, print the node
            // before calling itself for left subtree
            document.write(node.data + " ");
            printBoundaryLeft(node.left);
        }
        else if (node.right != null) {
            document.write(node.data + " ");
            printBoundaryLeft(node.right);
        }
   
        // do nothing if it is a leaf node, this way we avoid
        // duplicates in output
    }
   
    // A function to print all right boundary nodes,
    // except a leaf node
    // Print the nodes in BOTTOM UP manner
    function printBoundaryRight(node)
    {
        if (node == null)
            return;
   
        if (node.right != null) {
            // to ensure bottom up order, first call for right
            // subtree, then print this node
            printBoundaryRight(node.right);
            document.write(node.data + " ");
        }
        else if (node.left != null) {
            printBoundaryRight(node.left);
            document.write(node.data + " ");
        }
        // do nothing if it is a leaf node, this way we avoid
        // duplicates in output
    }
   
    // A function to do boundary traversal of a given binary tree
    function printBoundary(node)
    {
        if (node == null)
            return;
   
        document.write(node.data + " ");
   
        // Print the left boundary in top-down manner.
        printBoundaryLeft(node.left);
   
        // Print all leaf nodes
        printLeaves(node.left);
        printLeaves(node.right);
   
        // Print the right boundary in bottom-up manner
        printBoundaryRight(node.right);
    }
     
    root = new Node(20);
    root.left = new Node(8);
    root.left.left = new Node(4);
    root.left.right = new Node(12);
    root.left.right.left = new Node(10);
    root.left.right.right = new Node(14);
    root.right = new Node(22);
    root.right.right = new Node(25);
    printBoundary(root);
   
</script>

Output

20 8 4 10 14 25 22 

Time Complexity: O(n) where n is the number of nodes in binary tree.
Auxiliary Space: O(n)

Clean Code with returning the traversal:

[No direct printing + Iterative Version of the code]

Algorithm:

  1. Right Boundary – Go Right Right until no Right. Dont Include Leaf nodes. (as it leads to duplication)
  2. Left Boundary – Go Left Left until no Left. Dont Include Leaf nodes. (as it leads to duplication)
  3. Leaf Boundary – Do inorder/preorder, if leaf node add to the List.
  4. We pass the array as reference, so its the same memory location used by all functions, to coordinate the result at one place.

CODE:

C++




#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;
};
  
// Utility function to create a new tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = nullptr;
    return temp;
}
 
bool isLeaf(Node* node){
  if(node->left == NULL && node->right==NULL){
    return true;
  }
  return false;
}
 
void addLeftBound(Node * root, vector<int>& ans){
  //Go left left and no left then right but again check from left
  root = root->left;
  while(root){
    if(!isLeaf(root)) ans.push_back(root->data);
    if(root->left) root = root->left;
    else root = root->right;
  }
}
 
void addRightBound(Node * root, vector<int>& ans){
  //Go right right and no right then left but again check from right
  root = root->right;
  //As we need the reverse of this for Anticlockwise
  //refer above picture for better understanding
  stack<int> stk;
  while(root){
    if(!isLeaf(root)) stk.push(root->data);
    if(root->right) root = root->right;
    else root = root->left;
  }
  while(!stk.empty()){
    ans.push_back(stk.top());
    stk.pop();
  }
}
 
//its kind of preorder
void addLeaves(Node * root, vector<int>& ans){
  if(root==NULL){
    return;
  }
  if(isLeaf(root)){
    ans.push_back(root->data); //just store leaf nodes
    return;
  }
  addLeaves(root->left,ans);
  addLeaves(root->right,ans);
}
 
vector <int> boundary(Node *root)
{
  //Your code here
  vector<int> ans;
  if(root==NULL) return ans;
  if(!isLeaf(root)) ans.push_back(root->data); // if leaf then its done by addLeaf
  addLeftBound(root,ans);
  addLeaves(root,ans);
  addRightBound(root,ans);
 
  return ans;
 
}
 
int main()
{
    // Let us construct the tree given in the above diagram
    Node* root = newNode(20);
    root->left = newNode(8);
    root->left->left = newNode(4);
    root->left->right = newNode(12);
    root->left->right->left = newNode(10);
    root->left->right->right = newNode(14);
    root->right = newNode(22);
    root->right->right = newNode(25);
  
    vector<int>ans = boundary(root);
      for(int v : ans){
        cout<<v<<" ";
    }
      cout<<"\n";
  
    return 0;
}
 
//Code done by Balakrishnan R (rbkraj000)

Java




// Java program to print boundary traversal of binary tree
 
import java.io.*;
import java.util.*;
 
class BinaryTree {
    Node root;
    /* A binary tree node has data, pointer to left child
and a pointer to right child */
    static class Node {
        int data;
        Node left, right;
        Node(int d)
        {
            data = d;
            left = null;
            right = null;
        }
    }
 
    private boolean isLeaf(Node node)
    {
        if (node.left == null && node.right == null) {
            return true;
        }
        return false;
    }
 
    private void addLeftBound(Node root,
                              ArrayList<Integer> ans)
    {
        // Go left left and no left then right but again
        // check from left
        root = root.left;
        while (root != null) {
            if (!isLeaf(root)) {
                ans.add(root.data);
            }
 
            if (root.left != null) {
                root = root.left;
            }
            else {
                root = root.right;
            }
        }
    }
 
    private void addRightBound(Node root,
                               ArrayList<Integer> ans)
    {
        // Go right right and no right then left but again
        // check from right
        root = root.right;
        // As we need the reverse of this for Anticlockwise
        Stack<Integer> stk = new Stack<>();
        while (root != null) {
            if (!isLeaf(root)) {
                stk.push(root.data);
            }
            if (root.right != null) {
                root = root.right;
            }
            else {
                root = root.left;
            }
        }
 
        while (!stk.isEmpty()) {
            ans.add(stk.peek());
            stk.pop();
        }
    }
 
    // its kind of predorder
    private void addLeaves(Node root,
                           ArrayList<Integer> ans)
    {
        if (root == null) {
            return;
        }
 
        if (isLeaf(root)) {
            ans.add(root.data); // just store leaf nodes
            return;
        }
 
        addLeaves(root.left, ans);
        addLeaves(root.right, ans);
    }
 
    ArrayList<Integer> boundary(Node root)
    {
        ArrayList<Integer> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
 
        if (!isLeaf(root)) {
            ans.add(root.data); // if leaf then its done by
                                // addLeaves
        }
 
        addLeftBound(root, ans);
        addLeaves(root, ans);
        addRightBound(root, ans);
        return ans;
    }
 
    public static void main(String[] args)
    {
        // Let us construct the tree given in the above
        // diagram
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(20);
        tree.root.left = new Node(8);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(12);
        tree.root.left.right.left = new Node(10);
        tree.root.left.right.right = new Node(14);
        tree.root.right = new Node(22);
        tree.root.right.right = new Node(25);
 
        ArrayList<Integer> ans = tree.boundary(tree.root);
 
        for (int i = 0; i < ans.size(); i++) {
            System.out.print(ans.get(i) + " ");
        }
        System.out.println();
    }
}
 
// This code is contributed by Snigdha Patil

Python3




# Python program to print boundary traversal of binary tree
 
# A binary tree node has data, pointer to left child
# and a pointer to right child
class Node:
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
 
 
def isLeaf(node):
    if node.left == None and node.right == None:
        return True
    return False
 
 
def addLeftBound(root, res):
    # Go left left and no left then right but again check from left
    root = root.left
    while(root is not None):
        if isLeaf(root) is not True:
            res.append(root.data)
        if(root.left is not None):
            root = root.left
        else:
            root = root.right
 
 
def addRightBound(root, res):
    # Go right right and no right then left but again check from right
    root = root.right
    # As we need the reverse of this for Anticlockwise
    # refer above picture for better understanding
    stk = []
    while(root is not None):
        if isLeaf(root) is not True:
            stk.append(root.data)
        if root.right is not None:
            root = root.right
        else:
            root = root.left
 
    while(len(stk) != 0):
        res.append(stk.pop(-1))
 
         
# its kind of preorder
def addLeaves(root, res):
    if root is None:
        return
    if isLeaf(root) is True:
        res.append(root.data)  # just store leaf nodes
        return
    addLeaves(root.left, res)
    addLeaves(root.right, res)
 
 
def boundary(root, res):
    # Your code here
    if root is None:
        return
    if isLeaf(root) is not True:
        res.append(root.data)  # if leaf then its done by addLeaf
    addLeftBound(root, res)
    addLeaves(root, res)
    addRightBound(root, res)
 
 
if __name__ == '__main__':
    root = Node(20)
    root.left = Node(8)
    root.left.left = Node(4)
    root.left.right = Node(12)
    root.left.right.left = Node(10)
    root.left.right.right = Node(14)
    root.right = Node(22)
    root.right.right = Node(25)
 
    res = []
    boundary(root, res)
    for i in res:
        print(i)
 
# This code is contributed by Yash Agarwal(yashagarwal2852002)

C#




// C# program to print boundary traversal of binary tree
 
using System;
using System.Collections.Generic;
 
public class BinaryTree {
  class Node {
    public int data;
    public Node left, right;
 
    public Node(int val)
    {
      this.data = val;
      this.left = null;
      this.right = null;
    }
  }
 
  static bool isLeaf(Node node)
  {
    if (node.left == null && node.right == null) {
      return true;
    }
    return false;
  }
 
  static void addLeftBound(Node root, List<int> ans)
  {
    // Go left left and no left then right but again
    // check from left
    root = root.left;
    while (root != null) {
      if (!isLeaf(root))
        ans.Add(root.data);
      if (root.left != null)
        root = root.left;
      else
        root = root.right;
    }
  }
 
  static void addRightBound(Node root, List<int> ans)
  {
    // Go right right and no right then left but again
    // check from right
    root = root.right;
    Stack<int> stk = new Stack<int>();
    while (root != null) {
      if (!isLeaf(root))
        stk.Push(root.data);
      if (root.right != null)
        root = root.right;
      else
        root = root.left;
    }
    while (stk.Count != 0) {
      ans.Add(stk.Peek());
      stk.Pop();
    }
  }
 
  static void addLeaves(Node root, List<int> ans)
  {
    if (root == null)
      return;
    if (isLeaf(root)) {
      ans.Add(root.data);
      return;
    }
    addLeaves(root.left, ans);
    addLeaves(root.right, ans);
  }
 
  static void boundary(Node root, List<int> ans)
  {
    if (root == null)
      return;
    if (!isLeaf(root))
      ans.Add(root.data); // if leaf then its done by
    // addLeaf
    addLeftBound(root, ans);
    addLeaves(root, ans);
    addRightBound(root, ans);
  }
 
  static public void Main()
  {
    Node root = new Node(20);
    root.left = new Node(8);
    root.left.left = new Node(4);
    root.left.right = new Node(12);
    root.left.right.left = new Node(10);
    root.left.right.right = new Node(14);
    root.right = new Node(22);
    root.right.right = new Node(25);
 
    List<int> ans = new List<int>();
    boundary(root, ans);
    foreach(var i in ans) { Console.Write(i + " "); }
  }
}
 
// This code is contributed by Yash
// Agarwal(yashagarwal2852002)

Javascript




// JavaScript Program for the above approach
class Node{
    constructor(data){
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
function isLeaf(node){
    if(node.left == null && node.right == null) return true;
    return false;
}
 
function addLeftBound(root, ans)
{
 
    // Go left left and not left then right but again check from left
    root = root.left;
    while(root){
        if(!isLeaf(root)) ans.push(root.data);
        if(root.left) root = root.left;
        else root = root.right;
    }
}
 
function addRightBound(root, ans)
{
 
    // Go right right and no right then left but again check from right
    root = root.right;
     
    // As we need the reverse of this for Anticlockwise
    // refer above picture for better understanding
    let stk = [];
    while(root){
        if(!isLeaf(root)) stk.push(root.data);
        if(root.right) root = root.right;
        else root = root.left;
    }
    while(stk.length != 0){
        ans.push(stk.pop());
    }
}
 
// its kind of preorder
function addLeaves(root, ans){
    if(root == null) return;
    if(isLeaf(root)){
        ans.push(root.data); // just store leaf nodes
        return;
    }
    addLeaves(root.left, ans);
    addLeaves(root.right, ans);
}
 
function boundary(root)
{
 
    // Your Code here
    let ans = [];
    if(root == null) return ans;
    if(!isLeaf(root)) ans.push(root.data); // if leaf then its done by addLeaf
    addLeftBound(root,ans);
    addLeaves(root,ans);
    addRightBound(root,ans);
     
    return ans;
}
 
// Let us construct the tree given in the above diagram
let root = new Node(20);
root.left = new Node(8);
root.left.left = new Node(4);
root.left.right = new Node(12);
root.left.right.left = new Node(10);
root.left.right.right = new Node(14);
root.right = new Node(22);
root.right.right = new Node(25);
 
let ans = boundary(root);
for(let i = 0; i < ans.length; i++){
    console.log(ans[i] + " ");
}
 
// this code is contributed by Yash Agarwal(yashagarwal2852002)

Output

20 8 4 10 14 25 22 

Time Complexity: O(n) where n is the number of nodes in binary tree.
Auxiliary Space: O(n) 


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