Welcome to the daily solutions of our PROBLEM OF THE DAY (POTD). We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of Binary Tree but will also help you build up problem-solving skills.
POTD 22 November: Symmetric Tree:
Given a Binary Tree. Check whether it is Symmetric or not, i.e. whether the binary tree is a Mirror image of itself or not.
Example:
Input:
5
/ \
3 3
/ \ / \
8 9 9 8
Output:
True
Explanation:
Tree is mirror image of itself i.e. tree is symmetricInput:
5
/ \
8 7
\ \
4 3Output: False
Check for Symmetric Tree using Morris Traversal:
The idea is to traverse the tree using Morris Traversal and Reverse Morris Traversal to traverse the given binary tree and at each step check that the data of the current node is equal in both the traversals. If at any step the data of the nodes are different. Then, the given tree is not Symmetric Binary Tree.
Below is the implementation of the above approach:
class Solution {
public:
// return true/false denoting whether the tree is
// Symmetric or not
bool isSymmetric(struct Node* root)
{
// Code here
Node *curr1 = root, *curr2 = root;
// Loop to traverse the tree in
// Morris Traversal and
// Reverse Morris Traversal
while (curr1 != NULL && curr2 != NULL) {
if (curr1->left == NULL
&& curr2->right == NULL) {
if (curr1->data != curr2->data)
return false;
curr1 = curr1->right;
curr2 = curr2->left;
}
else if (curr1->left != NULL
&& curr2->right != NULL) {
Node* pre1 = curr1->left;
Node* pre2 = curr2->right;
while (pre1->right != NULL
&& pre1->right != curr1
&& pre2->left != NULL
&& pre2->left != curr2) {
pre1 = pre1->right;
pre2 = pre2->left;
}
if (pre1->right == NULL
&& pre2->left == NULL) {
// Here, we are threading the Node
pre1->right = curr1;
pre2->left = curr2;
curr1 = curr1->left;
curr2 = curr2->right;
}
else if (pre1->right == curr1
&& pre2->left == curr2) {
// Unthreading the nodes
pre1->right = NULL;
pre2->left = NULL;
if (curr1->data != curr2->data)
return false;
curr1 = curr1->right;
curr2 = curr2->left;
}
else
return false;
}
else
return false;
}
if (curr1 != curr2)
return false;
return true;
}
};
class GfG {
// return true/false denoting whether the tree is
// Symmetric or not
public static boolean isSymmetric(Node root)
{
// add your code here;
Node curr1 = root, curr2 = root;
// Loop to traverse the tree in
// Morris Traversal and
// Reverse Morris Traversal
while (curr1 != null && curr2 != null) {
if (curr1.left == null && curr2.right == null) {
if (curr1.data != curr2.data)
return false;
curr1 = curr1.right;
curr2 = curr2.left;
}
else if (curr1.left != null
&& curr2.right != null) {
Node pre1 = curr1.left;
Node pre2 = curr2.right;
while (pre1.right != null
&& pre1.right != curr1
&& pre2.left != null
&& pre2.left != curr2) {
pre1 = pre1.right;
pre2 = pre2.left;
}
if (pre1.right == null
&& pre2.left == null) {
// Here, we are threading the Node
pre1.right = curr1;
pre2.left = curr2;
curr1 = curr1.left;
curr2 = curr2.right;
}
else if (pre1.right == curr1
&& pre2.left == curr2) {
// Unthreading the nodes
pre1.right = null;
pre2.left = null;
if (curr1.data != curr2.data)
return false;
curr1 = curr1.right;
curr2 = curr2.left;
}
else
return false;
}
else
return false;
}
if (curr1 != curr2)
return false;
return true;
}
}
class Solution:
# return true/false denoting whether the tree is Symmetric or not
def isSymmetric(self, root):
# Your Code Here
curr1 = root
curr2 = root
# Loop to traverse the tree in
# Morris Traversal and
# Reverse Morris Traversal
while curr1 != None and curr2 != None:
if (curr1.left == None and
curr2.right == None):
if curr1.data != curr2.data:
return False
curr1 = curr1.right
curr2 = curr2.left
elif curr1.left != None and curr2.right != None:
pre1 = curr1.left
pre2 = curr2.right
while (pre1.right != None and
pre1.right != curr1 and
pre2.left != None and
pre2.left != curr2):
pre1 = pre1.right
pre2 = pre2.left
if pre1.right == None and pre2.left == None:
# Here, we are threading the Node
pre1.right = curr1
pre2.left = curr2
curr1 = curr1.left
curr2 = curr2.right
elif (pre1.right == curr1 and
pre2.left == curr2):
# Unthreading the nodes
pre1.right = None
pre2.left = None
if curr1.data != curr2.data:
return False
curr1 = curr1.right
curr2 = curr2.left
else:
return False
else:
return False
if curr1 != curr2:
return False
return True
Time Complexity: O(N), where N is the number of nodes in the binary tree.
Auxiliary Space: O(1)