Check if leaf traversal of two Binary Trees is same?

Leaf traversal is the sequence of leaves traversed from left to right. The problem is to check if the leaf traversals of two given Binary Trees are same or not.
Expected time complexity O(n). Expected auxiliary space O(h1 + h2) where h1 and h2 are heights of two Binary Trees.

Examples:

Input: Roots of below Binary Trees
         1            
        / \
       2   3      
      /   / \          
     4   6   7

          0
        /   \
       5     8      
        \   / \        
        4   6  7
Output: same
Leaf order traversal of both trees is 4 6 7     

Input: Roots of below Binary Trees
         0            
        / \
       1   2       
      / \       
     8   9   

         1
        / \
       4   3     
        \ / \        
        8 2  9

Output: Not Same
Leaf traversals of two trees are different.
For first, it is 8 9 2 and for second it is
8 2 9

We strongly recommend you to minimize your browser and try this yourself first.
A Simple Solution is to traverse the first tree and store leaves from left and right in an array. Then traverse other tree and store leaves in another array. Finally, compare two arrays. If both arrays are the same, then return true.

Below is the implementation of the above approach:- 

C++

// C++ code to check if leaf traversals
// of two Binary Trees are same or not.
#include <bits/stdc++.h>
using namespace std;

// Binary Tree Node
struct Node {
    int data;
    Node* left;
    Node* right;
};

// Returns new Node with data as
// input to below function.
Node* newNode(int d)
{
    Node* temp = new Node;
    temp->data = d;
    temp->left = NULL;
    temp->right = NULL;

    return temp;
}

// Function to store the leaf nodes of the tree
void collectLeafNodes(Node* root, vector<int>& storeLeaf)
{
    if (!root->left && !root->right) {
        storeLeaf.push_back(root->data);
    }
    if (root->left) {
        collectLeafNodes(root->left, storeLeaf);
    }
    if (root->right) {
        collectLeafNodes(root->right, storeLeaf);
    }
}

// iterative function.
// returns true if leaf traversals
// are same, else false.
bool isSame(Node* root1, Node* root2)
{
    vector<int> storeLeafTree1, storeLeafTree2;
    collectLeafNodes(root1, storeLeafTree1);
    collectLeafNodes(root2, storeLeafTree2);
    return storeLeafTree1 == storeLeafTree2;
}

// Driver Code
int main()
{
    Node* root1 = newNode(1);
    root1->left = newNode(2);
    root1->right = newNode(3);
    root1->left->left = newNode(4);
    root1->right->left = newNode(6);
    root1->right->right = newNode(7);

    Node* root2 = newNode(0);
    root2->left = newNode(1);
    root2->right = newNode(5);
    root2->left->right = newNode(4);
    root2->right->left = newNode(6);
    root2->right->right = newNode(7);

    if (isSame(root1, root2))
        cout << "Same";
    else
        cout << "Not Same";
    return 0;
}

// This code is contributed Aditya Kumar (adityakumar(129)

Java

// Java code for the above approach
import java.util.ArrayList;

class Node {
int data;
Node left, right;
  Node(int d) {
    data = d;
    left = right = null;
}
}

class BinaryTree {
Node newNode(int data) {
Node temp = new Node(data);
return temp;
}
  // Function to store the leaf nodes of the tree
  void collectLeafNodes(Node root, ArrayList<Integer> storeLeaf) {
    if (root.left == null && root.right == null) {
        storeLeaf.add(root.data);
    }
    if (root.left != null) {
        collectLeafNodes(root.left, storeLeaf);
    }
    if (root.right != null) {
        collectLeafNodes(root.right, storeLeaf);
    }
}

// iterative function.
// returns true if leaf traversals
// are same, else false.
boolean isSame(Node root1, Node root2) {
    ArrayList<Integer> storeLeafTree1 = new ArrayList<>();
    ArrayList<Integer> storeLeafTree2 = new ArrayList<>();
    collectLeafNodes(root1, storeLeafTree1);
    collectLeafNodes(root2, storeLeafTree2);
    return storeLeafTree1.equals(storeLeafTree2);
}

  // Driver Code
public static void main(String args[]) {
    BinaryTree bt = new BinaryTree();
    Node root1 = bt.newNode(1);
    root1.left = bt.newNode(2);
    root1.right = bt.newNode(3);
    root1.left.left = bt.newNode(4);
    root1.right.left = bt.newNode(6);
    root1.right.right = bt.newNode(7);

    Node root2 = bt.newNode(0);
    root2.left = bt.newNode(1);
    root2.right = bt.newNode(5);
    root2.left.right = bt.newNode(4);
    root2.right.left = bt.newNode(6);
    root2.right.right = bt.newNode(7);

    if (bt.isSame(root1, root2)) {
        System.out.println("Same");
    } else {
        System.out.println("Not Same");
    }
}
}

//This code is contributed by shivamsharma215

Python3

#Python code for the above approach
# Binary Tree Node
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# Returns new Node with data as
# input to below function.
def new_node(d):
    temp = Node(d)
    return temp

# Function to store the leaf nodes of the tree
def collect_leaf_nodes(root, store_leaf):
    if not root.left and not root.right:
        store_leaf.append(root.data)
    if root.left:
        collect_leaf_nodes(root.left, store_leaf)
    if root.right:
        collect_leaf_nodes(root.right, store_leaf)

# iterative function.
# returns true if leaf traversals
# are same, else false.
def is_same(root1, root2):
    store_leaf_tree1, store_leaf_tree2 = [], []
    collect_leaf_nodes(root1, store_leaf_tree1)
    collect_leaf_nodes(root2, store_leaf_tree2)
    return store_leaf_tree1 == store_leaf_tree2

# Driver Code
if __name__ == "__main__":
    root1 = new_node(1)
    root1.left = new_node(2)
    root1.right = new_node(3)
    root1.left.left = new_node(4)
    root1.right.left = new_node(6)
    root1.right.right = new_node(7)

    root2 = new_node(0)
    root2.left = new_node(1)
    root2.right = new_node(5)
    root2.left.right = new_node(4)
    root2.right.left = new_node(6)
    root2.right.right = new_node(7)

    if is_same(root1, root2):
        print("Same")
    else:
        print("Not Same")

C#

// C# code to check if leaf traversals
// of two Binary Trees are same or not.

using System;
using System.Linq;
using System.Collections.Generic;

class GFG
{
    // Binary Tree Node
    class Node {
        public int data;
        public Node left;
        public Node right;
        public Node(int data)
        {
            this.data = data;
            left = null;
            right = null;
        }
    }

    // Function to store the leaf nodes of the tree
    static void collectLeafNodes(Node root, List<int> storeLeaf)
    {
        if (root.left==null && root.right==null) {
            storeLeaf.Add(root.data);
        }
        if (root.left!=null) {
            collectLeafNodes(root.left, storeLeaf);
        }
        if (root.right!=null) {
            collectLeafNodes(root.right, storeLeaf);
        }
    }

    // iterative function.
    // returns true if leaf traversals
    // are same, else false.
    static bool isSame(Node root1, Node root2)
    {
        List<int> storeLeafTree1=new List<int>();
        List<int> storeLeafTree2=new List<int>();
        collectLeafNodes(root1, storeLeafTree1);
        collectLeafNodes(root2, storeLeafTree2);
        if(storeLeafTree1.Count!=storeLeafTree2.Count)
            return false;
        for(int i=0; i<storeLeafTree1.Count; i++)
            if(storeLeafTree1[i]!=storeLeafTree2[i])
                return false;

        return true;
    }

    // Driver Code
    static public void Main()
    {
        Node root1 = new Node(1);
        root1.left = new Node(2);
        root1.right = new Node(3);
        root1.left.left = new Node(4);
        root1.right.left = new Node(6);
        root1.right.right = new Node(7);

        Node root2 = new Node(0);
        root2.left = new Node(1);
        root2.right = new Node(5);
        root2.left.right = new Node(4);
        root2.right.left = new Node(6);
        root2.right.right = new Node(7);

        if (isSame(root1, root2))
            Console.Write("Same");
        else
            Console.Write("Not Same");
    }
}

Javascript

// Javascript code to check if leaf traversal
// of two binary trees are same or not.

// binary tree node
class Node{
    constructor(data){
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

// Returns new Node with data as
// input to below function.
function newNode(d){
    return new Node(d);
}

// Function to store the leaf nodes of the tree
function collectLeafNodes(root, storeLeaf){
    if(root.left == null && root.right == null){
        storeLeaf.push(root.data);
    }
    if(root.left != null){
        collectLeafNodes(root.left, storeLeaf);
    }
    if(root.right != null){
        collectLeafNodes(root.right, storeLeaf);
    }
}

// iterative function.
// returns true if leaf traversals
// are same, else false.
function isSame(root1, root2){
    let storeLeafTree1 = [];
    let storeLeafTree2 = [];
    collectLeafNodes(root1, storeLeafTree1);
    collectLeafNodes(root2, storeLeafTree2);
    return storeLeafTree1.join() == storeLeafTree2.join();
}

// driver code
let root1 = newNode(1);
root1.left = newNode(2);
root1.right = newNode(3);
root1.left.left = newNode(4);
root1.right.left = newNode(6);
root1.right.right = newNode(7);

let root2 = newNode(0);
root2.left = newNode(1);
root2.right = newNode(5);
root2.left.right = newNode(4);
root2.right.left = newNode(6);
root2.right.right = newNode(7);

if(isSame(root1, root2))
    console.log("Same");
else
    console.log("Not Same");

// this code is contributed by Yash Agarwal(yashagarwal2852002)

Output

Same

Complexity Analysis:

Time Complexity: O(m+n) where m and n are nodes in the first and second tree respectively. 
Space Complexity: O(m+n), where m and n are nodes in the first and second tree respectively. 

How to check with O(h1 + h2) space?
The idea is use iterative traversal. Traverse both trees simultaneously, look for a leaf node in both trees and compare the found leaves. All leaves must match.

Algorithm:

1. Create empty stacks stack1 and stack2 
   for iterative traversals of tree1 and tree2

2. insert (root of tree1) in stack1
   insert (root of tree2) in stack2

3. Stores current leaf nodes of tree1 and tree2
temp1 = (root of tree1) 
temp2 = (root of tree2)  

4. Traverse both trees using stacks
while (stack1 and stack2 parent empty) 
{
    // Means excess leaves in one tree
    if (if one of the stacks are empty)   
    return false

   // get next leaf node in tree1 
   temp1 = stack1.pop()
   while (temp1 is not leaf node) 
   {
        push right child to stack1     
    push left child to stack1
   }

   // get next leaf node in tree2     
   temp2 = stack2.pop()
   while (temp2 is not leaf node) 
   {
        push right child to stack2      
    push left child to stack2
   }

   // If leaves do not match return false
   if (temp1 != temp2)                  
       return false
}

5. If all leaves matched, return true

Below is the implementation of the above algorithm. 

C++

// C++ code to check if leaf traversals
// of two Binary Trees are same or not.
#include <bits/stdc++.h>
using namespace std;

// Binary Tree Node
struct Node {
    int data;
    Node* left;
    Node* right;
};

// Returns new Node with data as
// input to below function.
Node* newNode(int d)
{
    Node* temp = new Node;
    temp->data = d;
    temp->left = NULL;
    temp->right = NULL;

    return temp;
}

// checks if a given node is leaf or not.
bool isLeaf(Node* root)
{
    if (root == NULL)
        return false;
    if (!root->left && !root->right)
        return true;
    return false;
}

// iterative function.
// returns true if leaf traversals
// are same, else false.
bool isSame(Node* root1, Node* root2)
{
    stack<Node*> s1;
    stack<Node*> s2;

    // push root1 to empty stack s1.
    s1.push(root1);

    // push root2 to empty stack s2.
    s2.push(root2);

    // loop until either of stacks are non-empty.
    while (!s1.empty() || !s2.empty())
    {
        // this means one of the stacks has
        // extra leaves, hence return false.
        if (s1.empty() || s2.empty())
            return false;

        Node* temp1 = s1.top();
        s1.pop();
        while (temp1 != NULL && !isLeaf(temp1))
        {
            // Push right child if exists
            if (temp1->right)
                s1.push(temp1->right);

            // Push left child if exists
            if (temp1->left)
                s1.push(temp1->left);

            // Note that right child(if exists)
            // is pushed before left child(if exists).
            temp1 = s1.top();
            s1.pop();
        }

        Node* temp2 = s2.top();
        s2.pop();
        while (temp2 != NULL && !isLeaf(temp2))
        {
            // Push right child if exists
            if (temp2->right)
                s2.push(temp2->right);

            // Push left child if exists
            if (temp2->left)
                s2.push(temp2->left);
            temp2 = s2.top();
            s2.pop();
        }

        if (!temp1 && temp2 || temp1 && !temp2)
            return false;
        if (temp1 && temp2 && temp1->data!=temp2->data) {
            return false;
        }
    }

    // all leaves are matched
    return true;
}

// Driver Code
int main()
{
    Node* root1 = newNode(1);
    root1->left = newNode(2);
    root1->right = newNode(3);
    root1->left->left = newNode(4);
    root1->right->left = newNode(6);
    root1->right->right = newNode(7);

    Node* root2 = newNode(0);
    root2->left = newNode(1);
    root2->right = newNode(5);
    root2->left->right = newNode(4);
    root2->right->left = newNode(6);
    root2->right->right = newNode(7);

    if (isSame(root1, root2))
        cout << "Same";
    else
        cout << "Not Same";
    return 0;
}

// This code is contributed
// by AASTHA VARMA

Java

// Java program to check if two Leaf Traversal of
// Two Binary Trees is same or not
import java.util.*;
import java.lang.*;
import java.io.*;

// Binary Tree node
class Node
{
    int data;
    Node left, right;
    public Node(int x)
    {
        data = x;
        left = right = null;
    }
    public boolean isLeaf()
    {
        return (left == null && right == null);
    }
}

class LeafOrderTraversal
{
    // Returns true of leaf traversal of two trees is
    // same, else false
    public static boolean isSame(Node root1, Node root2)
    {
        // Create empty stacks.  These stacks are going
        // to be used for iterative traversals.
        Stack<Node> s1 = new Stack<Node>();
        Stack<Node> s2 = new Stack<Node>();

        s1.push(root1);
        s2.push(root2);

        // Loop until either of two stacks is not empty
        while (!s1.empty() || !s2.empty())
        {
            // If one of the stacks is empty means other
            // stack has extra leaves so return false
            if (s1.empty() || s2.empty())
                return false;

            Node temp1 = s1.pop();
            while (temp1 != null && !temp1.isLeaf())
            {
                // Push right and left children of temp1.
                // Note that right child is inserted
                // before left
                if (temp1.right != null)
                    s1.push(temp1.right);
                if (temp1.left != null)
                    s1.push(temp1.left);
                temp1 = s1.pop();
            }

            // same for tree2
            Node temp2 = s2.pop();
            while (temp2 != null && !temp2.isLeaf())
            {
                if (temp2.right != null)
                    s2.push(temp2.right);
                if (temp2.left != null)
                    s2.push(temp2.left);
                temp2 = s2.pop();
            }

            // If one is null and other is not, then
            // return false
            if (temp1 == null && temp2 != null)
                return false;
            if (temp1 != null && temp2 == null)
                return false;

            // If both are not null and data is not
            // same return false
            if (temp1 != null && temp2 != null)
            {
                if (temp1.data != temp2.data)
                    return false;
            }
        }

        // If control reaches this point, all leaves
        // are matched
        return true;
    }

    // Driver code
    public static void main(String[] args)
    {
        // Let us create trees in above example 1
        Node root1 = new Node(1);
        root1.left = new Node(2);
        root1.right = new Node(3);
        root1.left.left = new Node(4);
        root1.right.left = new Node(6);
        root1.right.right = new Node(7);

        Node root2 = new Node(0);
        root2.left = new Node(1);
        root2.right = new Node(5);
        root2.left.right = new Node(4);
        root2.right.left = new Node(6);
        root2.right.right = new Node(7);

        if (isSame(root1, root2))
            System.out.println("Same");
        else
            System.out.println("Not Same");
    }
}

Python3

# Python3 program to check if two Leaf
# Traversal of Two Binary Trees is same or not

# Binary Tree node


class Node:
    def __init__(self, x):
        self.data = x
        self.left = self.right = None

    def isLeaf(self):
        return (self.left == None and
                self.right == None)

# Returns true of leaf traversal of
# two trees is same, else false


def isSame(root1, root2):

    # Create empty stacks. These stacks are going
    # to be used for iterative traversals.
    s1 = []
    s2 = []

    s1.append(root1)
    s2.append(root2)

    # Loop until either of two stacks
    # is not empty
    while (len(s1) != 0 or len(s2) != 0):

        # If one of the stacks is empty means other
        # stack has extra leaves so return false
        if (len(s1) == 0 or len(s2) == 0):
            return False

        temp1 = s1.pop(-1)
        while (temp1 != None and not temp1.isLeaf()):

            # append right and left children of temp1.
            # Note that right child is inserted
            # before left
            if (temp1.right != None):
                s1.append(temp1. right)
            if (temp1.left != None):
                s1.append(temp1.left)
                temp1 = s1.pop(-1)

        # same for tree2
        temp2 = s2.pop(-1)
        while (temp2 != None and not temp2.isLeaf()):
            if (temp2.right != None):
                s2.append(temp2.right)
            if (temp2.left != None):
                s2.append(temp2.left)
            temp2 = s2.pop(-1)

        # If one is None and other is not,
        # then return false
        if (temp1 == None and temp2 != None):
            return False
        if (temp1 != None and temp2 == None):
            return False

        # If both are not None and data is
        # not same return false
        if (temp1 != None and temp2 != None):
            if (temp1.data != temp2.data):
                return False

    # If control reaches this point,
    # all leaves are matched
    return True


# Driver Code
if __name__ == '__main__':

    # Let us create trees in above example 1
    root1 = Node(1)
    root1.left = Node(2)
    root1.right = Node(3)
    root1.left.left = Node(4)
    root1.right.left = Node(6)
    root1.right.right = Node(7)

    root2 = Node(0)
    root2.left = Node(1)
    root2.right = Node(5)
    root2.left.right = Node(4)
    root2.right.left = Node(6)
    root2.right.right = Node(7)

    if (isSame(root1, root2)):
        print("Same")
    else:
        print("Not Same")

# This code is contributed by pranchalK

C#

// C# program to check if two Leaf Traversal
// of Two Binary Trees is same or not
using System;
using System.Collections.Generic;

// Binary Tree node
public class Node {
    public int data;
    public Node left, right;
    public Node(int x)
    {
        data = x;
        left = right = null;
    }
    public virtual bool Leaf
    {
        get {
          return (left == null && right == null);
        }
    }
}

class GFG {
    // Returns true of leaf traversal of
    // two trees is same, else false
    public static bool isSame(Node root1, Node root2)
    {
        // Create empty stacks. These stacks
        // are going to be used for iterative
        // traversals.
        Stack<Node> s1 = new Stack<Node>();
        Stack<Node> s2 = new Stack<Node>();

        s1.Push(root1);
        s2.Push(root2);

        // Loop until either of two stacks
        // is not empty
        while (s1.Count > 0 || s2.Count > 0)
        {
            // If one of the stacks is empty means other
            // stack has extra leaves so return false
            if (s1.Count == 0 || s2.Count == 0)
            {
                return false;
            }

            Node temp1 = s1.Pop();
            while (temp1 != null && !temp1.Leaf)
            {
                // Push right and left children of temp1.
                // Note that right child is inserted
                // before left
                if (temp1.right != null)
                {
                    s1.Push(temp1.right);
                }
                if (temp1.left != null)
                {
                    s1.Push(temp1.left);
                }
                temp1 = s1.Pop();
            }

            // same for tree2
            Node temp2 = s2.Pop();
            while (temp2 != null && !temp2.Leaf)
            {
                if (temp2.right != null)
                {
                    s2.Push(temp2.right);
                }
                if (temp2.left != null)
                {
                    s2.Push(temp2.left);
                }
                temp2 = s2.Pop();
            }

            // If one is null and other is not,
            // then return false
            if (temp1 == null && temp2 != null)
            {
                return false;
            }
            if (temp1 != null && temp2 == null)
            {
                return false;
            }

            // If both are not null and data
            // is not same return false
            if (temp1 != null && temp2 != null)
            {
                if (temp1.data != temp2.data)
                {
                    return false;
                }
            }
        }

        // If control reaches this point,
        // all leaves are matched
        return true;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        // Let us create trees in above example 1
        Node root1 = new Node(1);
        root1.left = new Node(2);
        root1.right = new Node(3);
        root1.left.left = new Node(4);
        root1.right.left = new Node(6);
        root1.right.right = new Node(7);

        Node root2 = new Node(0);
        root2.left = new Node(1);
        root2.right = new Node(5);
        root2.left.right = new Node(4);
        root2.right.left = new Node(6);
        root2.right.right = new Node(7);

        if (isSame(root1, root2)) {
            Console.WriteLine("Same");
        }
        else {
            Console.WriteLine("Not Same");
        }
    }
}

// This code is contributed by Shrikant13

Javascript

<script>

    // JavaScript program to check if two Leaf Traversal of
    // Two Binary Trees is same or not

    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }

    // Returns true of leaf traversal of two trees is
    // same, else false
    function isSame(root1, root2)
    {
        // Create empty stacks.  These stacks are going
        // to be used for iterative traversals.
        let s1 = [];
        let s2 = [];

        s1.push(root1);
        s2.push(root2);

        // Loop until either of two stacks is not empty
        while (s1.length > 0 || s2.length > 0)
        {
            // If one of the stacks is empty means other
            // stack has extra leaves so return false
            if (s1.length == 0 || s1.length == 0)
                return false;

            let temp1 = s1.pop();
            while (temp1 != null && !(temp1.left == null &&
            temp1.right == null))
            {
                // Push right and left children of temp1.
                // Note that right child is inserted
                // before left
                if (temp1.right != null)
                    s1.push(temp1.right);
                if (temp1.left != null)
                    s1.push(temp1.left);
                temp1 = s1.pop();
            }

            // same for tree2
            let temp2 = s2.pop();
            while (temp2 != null && !(temp2.left == null &&
            temp2.right == null))
            {
                if (temp2.right != null)
                    s2.push(temp2.right);
                if (temp2.left != null)
                    s2.push(temp2.left);
                temp2 = s2.pop();
            }

            // If one is null and other is not, then
            // return false
            if (temp1 == null && temp2 != null)
                return false;
            if (temp1 != null && temp2 == null)
                return false;

            // If both are not null and data is not
            // same return false
            if (temp1 != null && temp2 != null)
            {
                if (temp1.data != temp2.data)
                    return false;
            }
        }

        // If control reaches this point, all leaves
        // are matched
        return true;
    }

    // Let us create trees in above example 1
    let root1 = new Node(1);
    root1.left = new Node(2);
    root1.right = new Node(3);
    root1.left.left = new Node(4);
    root1.right.left = new Node(6);
    root1.right.right = new Node(7);

    let root2 = new Node(0);
    root2.left = new Node(1);
    root2.right = new Node(5);
    root2.left.right = new Node(4);
    root2.right.left = new Node(6);
    root2.right.right = new Node(7);

    if (isSame(root1, root2))
      document.write("Same");
    else
      document.write("Not Same");

</script>

Output

Same

Complexity Analysis:

Time Complexity: O(n)
Space Complexity: O(h), where h is max(h1,h2)

This article is contributed by Kumar Gaurav. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above