Mirror Tree | Practice | GeeksforGeeks

Given a Binary Tree, convert it into its mirror.
             
Example 1:
Input:
      1
    /  \
   2    3
Output: 3 1 2
Explanation: The tree is
   1    (mirror)  1
 /  \    =>      /  \
2    3          3    2
The inorder of mirror is 3 1 2

Example 2:
Input:
      10
     /  \
    20   30
   /  \
  40  60
Output: 30 10 60 20 40
Explanation: The tree is
      10               10
    /    \  (mirror) /    \
   20    30    =>   30    20
  /  \                   /   \
 40  60                 60   40
The inroder traversal of mirror is
30 10 60 20 40.
Your Task:
Just complete the function mirror() that takes node as paramter  and convert it into its mirror. The printing is done by the driver code only.
Expected Time Complexity: O(N).
Expected Auxiliary Space: O(Height of the Tree).
Constraints:
1 ≤ Number of nodes ≤ 105
1 ≤ Data of a node ≤ 105
0
amarrajsmart1972 weeks ago
C++ Solution.
if(node==NULL)       return ;       mirror(node->left);       mirror(node->right);          swap(node->left,node->right);       
0
hharshit81182 weeks ago
 void mirror(Node* node) {        if(node == NULL){          return;        }
   mirror(node->left);   mirror(node->right);           Node* temp = node->left;        node->left = node->right;        node->right = temp;      }
+1
harshsinghreal3 weeks ago
C++ 4LINES OF CODE
void mirror(Node* node) {


        if(node==NULL) return;
        swap(node->left,node->right);
        mirror(node->left);
        mirror(node->right);
        
        
    }
0
joyrockok3 weeks ago
class Solution {   // Function to convert a binary tree into its mirror tree.   void mirror(Node node) {       // Your code here       if(node == null)             return;        Node temp = node.left;       node.left = node.right;       node.right = temp;       mirror(node.left);      mirror(node.right);   }}
0
adityagagtiwari3 weeks ago
Just swap the links of left and right nodes.
Recursively do this for all the nodes. And voila.

class Solution {
   // Function to convert a binary tree into its mirror tree.
   void mirror(Node node) {
       // Your code here
       if(node==null)
       return;
       Node temp = node.left;
       node.left = node.right;
       node.right = temp;
       mirror(node.left);
       mirror(node.right);
   }
}
0
soumyadeepdatta344 weeks ago
class Solution {
    // Function to convert a binary tree into its mirror tree.
    void mirror(Node node) {
        // Your code here
        if (node == null || (node.left == null && node.right == null)) {
            return;
        } 
        mirror(node.left);
        mirror(node.right);
        Node tmp;
        tmp = node.left;
        node.left = node.right;
        node.right = tmp;
    }
}
0
harvey20014 weeks ago
void mirror(Node* root) {       if(root==NULL)       return;       if(root->left)       mirror(root->left);       if(root->right)       mirror(root->right);       swap(root->left,root->right);   }
0
vainalavinayvvk0981 month ago
class Solution:      def mirrortraversal(self,root,res):       if root:           root.left,root.right = root.right,root.left           self.mirrortraversal(root.left,res)           res.append(root.data)           self.mirrortraversal(root.right,res)
       return  res   #Function to convert a binary tree into its mirror tree.   def mirror(self,root):       # Code here       res = []       val = self.mirrortraversal(root,res)       return val
0
sharmaaditya130641 month ago
void mirror(Node* root) {
        if(!root)return;
       swap(root->left,root->right);
       mirror(root->left);
       mirror(root->right);
     
   }
0
crawler1 month ago
    void swap(Node *node){
        Node *left = node->left;
        node->left = node->right;
        node->right = left;
    }
    void mirror(Node* node) {
        if(!node){
            return;
        }
        mirror(node->left);
        mirror(node->right);
        swap(node);
    }
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