Given a Binary Tree, the task is to flatten it in order of ZigZag traversal of the tree. In the flattened binary tree, the left node of all the nodes must be NULL.
Examples:
Input:
1
/ \
5 2
/ \ / \
6 4 9 3
Output: 1 2 5 6 4 9 3
Input:
1
\
2
\
3
\
4
\
5
Output: 1 2 3 4 5
Approach: We will solve this problem by simulating the ZigZag traversal of Binary Tree.
Algorithm:
- Create two stacks, “c_lev” and “n_lev” and to store the nodes of current and next level Binary tree.
- Create a variable “prev” and initialise it by parent node.
- Push right and left children of parent in the c_lev stack.
- Apply ZigZag traversal. Lets say “curr” is top most element in “c_lev”. Then,
- If ‘curr’ is NULL, continue.
- Else push curr->left and curr->right on the stack “n_lev” in appropriate order. If we are performing left to right traversal then curr->left is pushed first else curr->right is pushed first.
- Set prev = curr.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Node of the binary tree struct node { int data; node* left; node* right; node(int data) { this->data = data; left = NULL; right = NULL; } }; // Function to flatten Binary tree void flatten(node* parent) { // Queue to store node // for BFS stack<node *> c_lev, n_lev; c_lev.push(parent->left); c_lev.push(parent->right); bool lev = 1; node* prev = parent; // Code for BFS while (c_lev.size()) { // Size of queue while (c_lev.size()) { // Front most node in // queue node* curr = c_lev.top(); c_lev.pop(); // Base case if (curr == NULL) continue; prev->right = curr; prev->left = NULL; prev = curr; // Pushing new elements // in queue if (!lev) n_lev.push(curr->left); n_lev.push(curr->right); if (lev) n_lev.push(curr->left); } lev = (!lev); c_lev = n_lev; while (n_lev.size()) n_lev.pop(); } prev->left = NULL; prev->right = NULL; } // Function to print flattened // binary tree void print(node* parent) { node* curr = parent; while (curr != NULL) cout << curr->data << " ", curr = curr->right; } // Driver code int main() { node* root = new node(1); root->left = new node(5); root->right = new node(2); root->left->left = new node(6); root->left->right = new node(4); root->right->left = new node(9); root->right->right = new node(3); // Calling required functions flatten(root); print(root); return 0; } |
1 2 5 6 4 9 3
Time Complexity: O(N)
Space Complexity: O(N) where N is the size of Binary Tree.
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