Postorder Traversal of Binary Tree
Postorder traversal is defined as a type of tree traversal which follows the Left-Right-Root policy such that for each node:
- The left subtree is traversed first
- Then the right subtree is traversed
- Finally, the root node of the subtree is traversed

Postorder traversal
Algorithm for Postorder Traversal of Binary Tree:
The algorithm for postorder traversal is shown as follows:
Postorder(root):
- Follow step 2 to 4 until root != NULL
- Postorder (root -> left)
- Postorder (root -> right)
- Write root -> data
- End loop
How does Postorder Traversal of Binary Tree Work?
Consider the following tree:

Example of Binary Tree
If we perform a postorder traversal in this binary tree, then the traversal will be as follows:
Step 1: The traversal will go from 1 to its left subtree i.e., 2, then from 2 to its left subtree root, i.e., 4. Now 4 has no subtree, so it will be visited.
Node 4 is visited
Step 2: As the left subtree of 2 is visited completely, now it will traverse the right subtree of 2 i.e., it will move to 5. As there is no subtree of 5, it will be visited.
Node 5 is visited
Step 3: Now both the left and right subtrees of node 2 are visited. So now visit node 2 itself.
Node 2 is visited
Step 4: As the left subtree of node 1 is traversed, it will now move to the right subtree root, i.e., 3. Node 3 does not have any left subtree, so it will traverse the right subtree i.e., 6. Node 6 has no subtree and so it is visited.
Node 6 is visited
Step 5: All the subtrees of node 3 are traversed. So now node 3 is visited.
Node 3 is visited
Step 6: As all the subtrees of node 1 are traversed, now it is time for node 1 to be visited and the traversal ends after that as the whole tree is traversed.
The complete tree is visited
So the order of traversal of nodes is 4 -> 5 -> 2 -> 6 -> 3 -> 1.
Program to implement Postorder Traversal of Binary Tree
Below is the code implementation of the postorder traversal:
C++
// C++ program for postorder traversals #include <bits/stdc++.h> using namespace std; // Structure of a Binary Tree Node struct Node { int data; struct Node *left, *right; Node( int v) { data = v; left = right = NULL; } }; // Function to print postorder traversal void printPostorder( struct Node* node) { if (node == NULL) return ; // First recur on left subtree printPostorder(node->left); // Then recur on right subtree printPostorder(node->right); // Now deal with the node cout << node->data << " " ; } // Driver code int main() { struct Node* root = new Node(1); root->left = new Node(2); root->right = new Node(3); root->left->left = new Node(4); root->left->right = new Node(5); root->right->right = new Node(6); // Function call cout << "Postorder traversal of binary tree is: \n" ; printPostorder(root); return 0; } |
Java
// Java program for postorder traversals import java.util.*; // Structure of a Binary Tree Node class Node { int data; Node left, right; Node( int v) { data = v; left = right = null ; } } class GFG { // Function to print postorder traversal static void printPostorder(Node node) { if (node == null ) return ; // First recur on left subtree printPostorder(node.left); // Then recur on right subtree printPostorder(node.right); // Now deal with the node System.out.print(node.data + " " ); } // Driver code public static void main(String[] args) { Node root = new Node( 1 ); root.left = new Node( 2 ); root.right = new Node( 3 ); root.left.left = new Node( 4 ); root.left.right = new Node( 5 ); root.right.right = new Node( 6 ); // Function call System.out.println( "Postorder traversal of binary tree is: " ); printPostorder(root); } } // This code is contributed by prasad264 |
Python3
# Python program for postorder traversals # Structure of a Binary Tree Node class Node: def __init__( self , v): self .data = v self .left = None self .right = None # Function to print postorder traversal def printPostorder(node): if node = = None : return # First recur on left subtree printPostorder(node.left) # Then recur on right subtree printPostorder(node.right) # Now deal with the node print (node.data, end = ' ' ) # Driver code if __name__ = = '__main__' : root = Node( 1 ) root.left = Node( 2 ) root.right = Node( 3 ) root.left.left = Node( 4 ) root.left.right = Node( 5 ) root.right.right = Node( 6 ) # Function call print ( "Postorder traversal of binary tree is:" ) printPostorder(root) |
C#
// C# program for postorder traversals using System; // Structure of a Binary Tree Node public class Node { public int data; public Node left, right; public Node( int v) { data = v; left = right = null ; } } public class GFG { // Function to print postorder traversal static void printPostorder(Node node) { if (node == null ) return ; // First recur on left subtree printPostorder(node.left); // Then recur on right subtree printPostorder(node.right); // Now deal with the node Console.Write(node.data + " " ); } static public void Main() { // Code Node root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); root.right.right = new Node(6); // Function call Console.WriteLine( "Postorder traversal of binary tree is: " ); printPostorder(root); } } // This code is contributed by karthik. |
Javascript
// Structure of a Binary Tree Node class Node { constructor(v) { this .data = v; this .left = null ; this .right = null ; } } // Function to print postorder traversal function printPostorder(node) { if (node == null ) { return ; } // First recur on left subtree printPostorder(node.left); // Then recur on right subtree printPostorder(node.right); // Now deal with the node console.log(node.data + " " ); } // Driver code function main() { let root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); root.right.right = new Node(6); // Function call console.log( "Postorder traversal of binary tree is: \n" ); printPostorder(root); } main(); |
Postorder traversal of binary tree is: 4 5 2 6 3 1
Explanation:

How postorder traversal works
Complexity Analysis:
Time Complexity: O(N) where N is the total number of nodes. Because it traverses all the nodes at least once.
Auxiliary Space: O(1) if no recursion stack space is considered. Otherwise, O(h) where h is the height of the tree
- In the worst case, h can be the same as N (when the tree is a skewed tree)
- In the best case, h can be the same as logN (when the tree is a complete tree)
Use cases of Postorder Traversal:
Some use cases of postorder traversal are:
- This is used for tree deletion.
- It is also useful to get the postfix expression from an expression tree.
Related articles:
- Types of Tree traversals
- Iterative Postorder traversal (using two stacks)
- Iterative Postorder traversal (using one stack)
- Postorder of Binary Tree without recursion and without stack
- Find Postorder traversal of BST from preorder traversal
- Morris traversal for Postorder
- Print postorder traversal from preoreder and inorder traversal
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