Given a binary tree, the task is to flatten it in order of its post-order traversal. In the flattened binary tree, the left node of all the nodes must be NULL.
Examples:
Input:
5
/ \
3 7
/ \ / \
2 4 6 8
Output: 2 4 3 6 8 7 5
Input:
1
\
2
\
3
\
4
\
5
Output: 5 4 3 2 1
A simple approach will be to recreate the Binary Tree from its post-order traversal. This will take O(N) extra space were N is the number of nodes in BST.
A better solution is to simulate post-order traversal of the given binary tree.
- Create a dummy node.
- Create variable called ‘prev’ and make it point to the dummy node.
- Perform post-order traversal and at each step.
- Set prev -> right = curr
- Set prev -> left = NULL
- Set prev = curr
This will improve the space complexity to O(H) in the worst case as post-order traversal takes O(H) extra space.
Below is the implementation of the above approach:
C++
// 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 print the flattened // binary Tree void print(node* parent) { node* curr = parent; while (curr != NULL) cout << curr->data << " ", curr = curr->right; } // Function to perform post-order traversal // recursively void postorder(node* curr, node*& prev) { // Base case if (curr == NULL) return; postorder(curr->left, prev); postorder(curr->right, prev); prev->left = NULL; prev->right = curr; prev = curr; } // Function to flatten the given binary tree // using post order traversal node* flatten(node* parent) { // Dummy node node* dummy = new node(-1); // Pointer to previous element node* prev = dummy; // Calling post-order traversal postorder(parent, prev); prev->left = NULL; prev->right = NULL; node* ret = dummy->right; // Delete dummy node delete dummy; return ret; } // Driver code int main() { node* root = new node(5); root->left = new node(3); root->right = new node(7); root->left->left = new node(2); root->left->right = new node(4); root->right->left = new node(6); root->right->right = new node(8); print(flatten(root)); return 0; } |
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Java
// Java implementation of the approach class GFG { // Node of the binary tree static class node { int data; node left; node right; node(int data) { this.data = data; left = null; right = null; } }; static node prev; // Function to print the flattened // binary Tree static void print(node parent) { node curr = parent; while (curr != null) { System.out.print(curr.data + " "); curr = curr.right; } } // Function to perform post-order traversal // recursively static void postorder(node curr) { // Base case if (curr == null) return; postorder(curr.left); postorder(curr.right); prev.left = null; prev.right = curr; prev = curr; } // Function to flatten the given binary tree // using post order traversal static node flatten(node parent) { // Dummy node node dummy = new node(-1); // Pointer to previous element prev = dummy; // Calling post-order traversal postorder(parent); prev.left = null; prev.right = null; node ret = dummy.right; // Delete dummy node dummy = null; return ret; } // Driver code public static void main(String[] args) { node root = new node(5); root.left = new node(3); root.right = new node(7); root.left.left = new node(2); root.left.right = new node(4); root.right.left = new node(6); root.right.right = new node(8); print(flatten(root)); } } // This code is contributed by PrinciRaj1992 |
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Python
# Python implementation of above algorithm # Utility class to create a node class node: def __init__(self, key): self.data = key self.left = self.right = None # Function to print the flattened # binary Tree def print_(parent): curr = parent while (curr != None): print( curr.data ,end = " ") curr = curr.right prev = None # Function to perform post-order traversal # recursively def postorder( curr ): global prev # Base case if (curr == None): return postorder(curr.left) postorder(curr.right) prev.left = None prev.right = curr prev = curr # Function to flatten the given binary tree # using post order traversal def flatten(parent): global prev # Dummy node dummy = node(-1) # Pointer to previous element prev = dummy # Calling post-order traversal postorder(parent) prev.left = None prev.right = None ret = dummy.right return ret # Driver code root = node(5) root.left = node(3) root.right = node(7) root.left.left = node(2) root.left.right = node(4) root.right.left = node(6) root.right.right = node(8) print_(flatten(root)) # This code is contributed by Arnab Kundu |
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C#
// C# implementation of the approach using System; class GFG { // Node of the binary tree public class node { public int data; public node left; public node right; public node(int data) { this.data = data; left = null; right = null; } }; static node prev; // Function to print the flattened // binary Tree static void print(node parent) { node curr = parent; while (curr != null) { Console.Write(curr.data + " "); curr = curr.right; } } // Function to perform post-order traversal // recursively static void postorder(node curr) { // Base case if (curr == null) return; postorder(curr.left); postorder(curr.right); prev.left = null; prev.right = curr; prev = curr; } // Function to flatten the given binary tree // using post order traversal static node flatten(node parent) { // Dummy node node dummy = new node(-1); // Pointer to previous element prev = dummy; // Calling post-order traversal postorder(parent); prev.left = null; prev.right = null; node ret = dummy.right; // Delete dummy node dummy = null; return ret; } // Driver code public static void Main(String[] args) { node root = new node(5); root.left = new node(3); root.right = new node(7); root.left.left = new node(2); root.left.right = new node(4); root.right.left = new node(6); root.right.right = new node(8); print(flatten(root)); } } // This code is contributed by Princi Singh |
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Output:
2 4 3 6 8 7 5
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