Check if leaf traversal of two Binary Trees is same?

Leaf traversal is sequence of leaves traversed from left to right. The problem is to check if 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 traverse 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 same, then return true.

The above solution requires O(m+n) extra space where m and n are nodes in 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 Java implementation of above algorithm.

```// 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 program to test
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");
}
}
```

Output:

`Same`

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