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Mirror of n-ary Tree in C++
Problem statement
Given a Tree where every node contains variable number of children, convert the tree to its mirror
Example
If n-ary tree is −
Then it’s mirror is −
Example
#include <bits/stdc++.h>
using namespace std;
struct node {
int data;
vector<node *>child;
};
node *newNode(int x) {
node *temp = new node;
temp->data = x;
return temp;
}
void mirrorTree(node * root) {
if (root == NULL) {
return;
}
int n = root->child.size();
if (n < 2) {
return;
}
for (int i = 0; i < n; i++) {
mirrorTree(root->child[i]);
}
reverse(root->child.begin(), root->child.end());
}
void printTree(node * root) {
if (root == NULL) {
return;
}
queue<node *>q;
q.push(root);
int level = 0;
while (!q.empty()) {
int n = q.size();
++level;
cout << "Level " << level << ": ";
while (n > 0) {
node * p = q.front();
q.pop();
cout << p->data << " ";
for (int i=0; i<p->child.size(); i++) {
q.push(p->child[i]);
}
n--;
}
cout << endl;
}
}
int main() {
node *root = newNode(20);
(root->child).push_back(newNode(10));
(root->child).push_back(newNode(15));
(root->child[0]->child).push_back(newNode(1));
(root->child[0]->child).push_back(newNode(2));
(root->child[0]->child).push_back(newNode(3));
(root->child[1]->child).push_back(newNode(4));
cout << "Tree traversal before mirroring\n";
printTree(root);
mirrorTree(root);
cout << "\nTree traversal after mirroring\n";
printTree(root);
return 0;
}When you compile and execute above program. It generates following output −
Output
Tree traversal before mirroring Level 1: 20 Level 2: 10 15 Level 3: 1 2 3 4 Tree traversal after mirroring Level 1: 20 Level 2: 15 10 Level 3: 4 3 2 1
Updated on: 20-Dec-2019
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