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Symmetric Tree

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Average time to solve is 20m
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OLX GroupCIS - Cyber InfrastructureIBM

Problem statement

You are given a binary tree, where the data present in each node is an integer. You have to find whether the given tree is symmetric or not.

Symmetric tree is a binary tree, whose mirror image is exactly the same as the original tree.

For Example:

sym_tree

Detailed explanation ( Input/output format, Notes, Images )
Input format:
The only line of input contains the binary tree node elements in the level order form. The values of nodes are separated by a single space in a single line. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted in the below image would be :

alt text

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation :

Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null (-1).
Note :
The above format was just to provide clarity on how the input is formed for a given tree. 

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
Output format:
The output consists of a single line containing "Symmetric" if the given tree is symmetric, else "Asymmetric".
Note:
You are not required to print the expected output; it has already been taken care of, Just implement the function.
Constraints:
0 <= N <= 10^5
1 <= Data <= 10^5

Where 'N' denotes the number of nodes in the given binary tree and 'Data' denotes the node value.

Time limit: 1sec
Sample Input 1:
1 2 2 3 4 4 3 -1 -1 -1 -1 -1 -1 -1 -1
Sample Output 1:
Symmetric
Explanation For Sample 1:
This is a symmetric tree:

sym_tree

Sample Input 2:
1 2 3 4 -1 -1 -1 -1 -1
Sample Output 2:
Asymmetric
Explanation For Sample 2:
This is an asymmetric tree:

sym_tree

Hint

Think about how you can obtain a mirror image of the tree. Does it have something to do with how you traverse the tree? More specifically, in case of a binary tree, which child do you traverse first?

Approaches (1)
Inorder And Reverse Inorder

We do two inorder traversals recursively of the tree simultaneously. For the first traversal, we traverse the left child first and then the right child, and in the second traversal, we traverse the right child first and then the left child. Let cur1 and cur2 be the current nodes of the first and second traversals respectively and initially both of them are passed as root. Then the algorithm follows:

 

  • If cur1 and cur2 both are NULL, we terminate the algorithm by returning true. (This is the base case)
  • If one of these nodes is NULL, and the other is non-NULL, we terminate the algorithm by returning false.
  • If we have come this far, it means that both nodes are non-NULL. We check if the data in these two nodes are equal. If not, we terminate and return false.
  • Uptil this it is true, now we recurse for the left child of cur1 and right child of cur2, as well as the right child of cur1 and the left child of cur2.
Time Complexity

O(N), Where N is the number of nodes present in the tree

 

Since we traverse every node once, the time complexity is O(N).

Space Complexity

O(H), where H is the height of the tree (For a balanced tree, H = log(N))

 

This is because the recursion stack has a maximum depth of the height of our tree.

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Symmetric Tree
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Interview problems

if you're java and getting 1 failed test case

use `if (!root1.data.equals(root2.data))` instead of `if (!root1.data ==root2.data)) {`

 

because treenode contains Integer object not primitive int. so equals operator fails for >127 values with Integer objects

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2 upvotes

Interview problems

easy to understand ,beats 99.6%solutions in c++

bool symmetric(BinaryTreeNode<int>*root1,BinaryTreeNode<int>* root2){

    if(root1==nullptr && root2==nullptr){

        return true;

    }

    if(root1==nullptr && root2!=nullptr){

        return false;

    }

    if(root2==nullptr && root1!=nullptr){

        return false;

    }

    if(root1->data!=root2->data){

        return false;

    }

    bool a=symmetric(root1->left,root2->right);

    bool b=symmetric(root1->right,root2->left);

    return a==b;

}

 

bool isSymmetric(BinaryTreeNode<int>* root)

{

    if(root==nullptr){

        return true;

    }   

    return symmetric(root->left,root->right);

}

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Interview problems

why your code is failing on one testcase (root is null) here's the solution

Just return true if the root is NULL 


bool isSameTree(BinaryTreeNode<int>* p, BinaryTreeNode<int>* q) {
    if (p == NULL || q == NULL) {
        return p == q;
    }
    return (p->data == q->data) && isSameTree(p->left, q->right) && isSameTree(p->right, q->left);
}

bool isSymmetric(BinaryTreeNode<int>* root) {
    if (root == NULL) {
        return true;
    }
    return isSameTree(root->left, root->right);
}

datastructures

Tree Traversals

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1 upvote

Interview problems

Symmetric Tree Easy CPP Solution 100 %

void left(BinaryTreeNode<int>* root,vector<int>& ans1)

 

{

    if(root == NULL)

        return ;

    ans1.push_back(root->data);

    left(root->left,ans1);

    left(root->right,ans1);

}

 

void right(BinaryTreeNode<int>* root,vector<int>& ans2)

{

    if(root == NULL)

        return ;

    ans2.push_back(root->data);

    right(root->right,ans2);

    right(root->left,ans2);

}

 

bool isSymmetric(BinaryTreeNode<int>* root)

{

    vector<int> ans1,ans2;

    left(root,ans1);

    right(root,ans2);

    if(ans1 == ans2)

        return true;

    return false; 

}

 

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0 upvotes

Interview problems

Easy solution with c++ | Inorder traversal

void left(BinaryTreeNode<int>* root,vector<int>& ans1)

{

    if(root == NULL)

        return ;

    ans1.push_back(root->data);

    left(root->left,ans1);

    left(root->right,ans1);

}

void right(BinaryTreeNode<int>* root,vector<int>& ans2)

{

    if(root == NULL)

        return ;

    ans2.push_back(root->data);

    right(root->right,ans2);

    right(root->left,ans2);

}

 

bool isSymmetric(BinaryTreeNode<int>* root)

{

    vector<int> ans1,ans2;

    left(root,ans1);

    right(root,ans2);

 

    if(ans1 == ans2)

        return true;

    return false; 

}

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Interview problems

Java Code | Handles all the cases successfully| Easy to understand

public class Solution {
     private  static boolean isSymmetricUtil(BinaryTreeNode<Integer> root1, BinaryTreeNode<Integer> root2) {
       if (root1 == null && root2 == null) {
            return true;
        }
        if (root1 == null || root2 == null) {
            return false;
        }
        if (!root1.data.equals(root2.data)) {
            return false;
        }
        else return isSymmetricUtil(root1.left,root2.right) && isSymmetricUtil(root1.right,root2.left);
    }
    public static boolean isSymmetric(BinaryTreeNode<Integer> root) {
        
         if(root==null)return true;
         return isSymmetricUtil(root.left,root.right);
    }
}
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0 upvotes

Interview problems

Why you are failing in one test case even though the same code works on LeetCode (EXPLAINED)

So even i was wondering as to why i was not able to pass one test case for this case even though the same code was working in Leetcode. The reason i can think of is that we need to use .equals() rather == because .equals() will check the value of the two objects we are comparing whereas in == operator checks for reference equality.

Below is the Code:
public class Solution {
public static boolean isMirror(BinaryTreeNode<Integer>t1, BinaryTreeNode<Integer> t2){
    if(t1==null && t2==null) return true;
    if(t1==null || t2==null) return false;
    return (t1.data.equals(t2.data)) && isMirror(t1.left,t2.right) && isMirror(t1.right,t2.left);
}
    public static boolean isSymmetric(BinaryTreeNode<Integer> root) {
        if(root==null) return true;
        return isMirror(root,root);
    }
}

java

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1 upvote

Interview problems

Java Solution is working in GFG and Leetcode but here one Test case is failing could you please share the failed test case

public class Solution {

 public static boolean isSymmetric(BinaryTreeNode<Integer> root) {

        // Write your code here.

        if(root == null) {

            return true;

        }

        return isIdentical(root.left, root.right);

    }

 

    public static boolean isIdentical(BinaryTreeNode<Integer> root1, BinaryTreeNode<Integer> root2) {

        if(root1 == null && root2 == null) {

            return true;

        }

        else if(root1 == null && root2 != null) {

            return false;

        }

        else if(root1 != null && root2 == null) {

            return false;

        }

        else {

            return ((root1.data == root2.data)

                        && isIdentical(root1.left, root2.right)

                        && isIdentical(root1.right, root2.left)

                    );

        }

    }

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Interview problems

Striver's easy recursive solution in C++ | Symmetric tree

my github : https://github.com/yashswag22

bool isSymmetricHelp(BinaryTreeNode<int>* left, BinaryTreeNode<int>* right){
    if(left == NULL || right == NULL)
    return left == right;

    if(left->data != right->data) return false;

    return isSymmetricHelp(left->left,right->right) && isSymmetricHelp(left->right,right->left);
}

bool isSymmetric(BinaryTreeNode<int>* root){
    return root == NULL || isSymmetricHelp(root->left, root->right);
}

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Interview problems

how to solve symmetric tree question , solution in c++

#include<bits/stdc++.h>
// void printMap(const map<int, vector<pair<int, int>>>& nodes) {
//     for (const auto& lvl_pair : nodes) {
//         int lvl = lvl_pair.first;
//         const vector<pair<int, int>>& vec = lvl_pair.second;
        
//         cout << "{" << lvl;
//         for (const auto& p : vec) {
//             cout << ", {" << p.first << ", " << p.second << "}";
//         }
//         cout << "}" << endl;
//     }
// }

bool isSymmetric(BinaryTreeNode<int>* root)
{
   if(root == NULL) return true;
 
    if(root->left == NULL && root->right == NULL) return true;

      map <int,vector<pair<int,int>>> nodes;
   queue<pair<int,pair<int,BinaryTreeNode<int>*>>> q;
            //y,x
   //q.push({0,{0,root}});

   if(root->left)
   q.push({1,{-1,root->left}});

   if(root->right)
   q.push({1,{1,root->right}});


   while(!q.empty()){
       pair<int,pair<int,BinaryTreeNode<int>*>> temp = q.front();
       q.pop();
       int lvl = temp.first;
       int x = temp.second.first;
       BinaryTreeNode<int>* frontnode = temp.second.second;

        if(frontnode != NULL)
        nodes[lvl].push_back({x,frontnode->data});

        if(frontnode->left)
        q.push({lvl+1,{x-1,frontnode->left}});

        if(frontnode->right)
         q.push({lvl+1,{x+1,frontnode->right}});

   }

    //printMap(nodes);

   for(auto i:nodes){ // give <lvl,vector<pair<int,int>>>
       // vector<pair<int,int>> temp = i.second;
       int siz = i.second.size();
        if(siz%2 != 0) return false;
        for(int j=0;j<siz/2;j++){
            pair<int,int> temp = i.second[j];
            pair<int,int> temp2 = i.second[(siz-1)-j];

            if(temp.first != -1*(temp2.first)) return false;
            else{
                if(temp.second != temp2.second) return false;
            }

            // cout<<temp.first<<" "<<temp.second<<" ";
            // cout<<temp2.first<<" "<<temp2.second;
            
        }
       // cout<<endl;

   }

   return true;
}

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