Tree Traversal Postorder Left Right Root in DSA C++

Mental Model

Visit the left child, then the right child, and finally the node itself. This order ensures children are processed before their parent.

Analogy: Imagine cleaning a tree house: first clean the left room, then the right room, and only after both are clean, clean the main hall.

    1
   / \
  2   3
 / \
4   5

↑ root at 1

Dry Run Walkthrough

Input: Tree: 1 with left child 2 and right child 3; 2 has children 4 (left) and 5 (right)

Goal: Print nodes in postorder: left subtree, right subtree, then root

Step 1: Go to left child of 1 (node 2)

1 -> [curr->2] -> 3
2 -> 4 -> 5
↑ at node 2

Why: Start with left subtree to follow postorder

Step 2: Go to left child of 2 (node 4)

1 -> 2 -> 3
2 -> [curr->4] -> 5
↑ at node 4

Why: Keep going left to reach leftmost node

Step 3: Node 4 has no children, print 4

Printed: 4
Back to node 2

Why: Left subtree done, print leaf node

Step 4: Go to right child of 2 (node 5)

1 -> 2 -> 3
2 -> 4 -> [curr->5]
↑ at node 5

Why: Now process right subtree of node 2

Step 5: Node 5 has no children, print 5

Printed: 4 5
Back to node 2

Why: Right subtree done, print leaf node

Step 6: Print node 2 after its children

Printed: 4 5 2
Back to node 1

Why: Both children processed, print parent

Step 7: Go to right child of 1 (node 3)

1 -> [curr->3]
↑ at node 3

Why: Process right subtree of root

Step 8: Node 3 has no children, print 3

Printed: 4 5 2 3
Back to node 1

Why: Right subtree done, print leaf node

Step 9: Print root node 1 last

Printed: 4 5 2 3 1

Why: Root printed after all children

Result:

4 5 2 3 1

Annotated Code

DSA C++

#include <iostream>
using namespace std;

struct Node {
    int val;
    Node* left;
    Node* right;
    Node(int v) : val(v), left(nullptr), right(nullptr) {}
};

void postorder(Node* root) {
    if (root == nullptr) return; // base case: empty node
    postorder(root->left); // visit left subtree
    postorder(root->right); // visit right subtree
    cout << root->val << " "; // visit root last
}

int main() {
    // Build tree:
    //     1
    //    / \
    //   2   3
    //  / \
    // 4   5
    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);

    postorder(root);
    cout << endl;
    return 0;
}

if (root == nullptr) return;

stop recursion at empty node

postorder(root->left);

traverse left subtree first

postorder(root->right);

then traverse right subtree

cout << root->val << " ";

print current node after children

Output

4 5 2 3 1 

Complexity Analysis

Time: O(n) because each node is visited exactly once

Space: O(h) where h is tree height due to recursion stack

vs Alternative: Compared to iterative traversal, recursive postorder is simpler but uses call stack; iterative uses explicit stack but can be more complex

Edge Cases

Empty tree (root is nullptr)

Function returns immediately without printing

DSA C++

if (root == nullptr) return;

Single node tree

Prints the single node value

DSA C++

if (root == nullptr) return;

Tree with only left children

Visits all left nodes in order, printing from bottom up

DSA C++

postorder(root->left);

When to Use This Pattern

When you need to process children before their parent, especially in cleanup or evaluation problems, use postorder traversal.

Common Mistakes

Mistake: Printing the root before children (preorder) instead of after
Fix: Move the print statement after recursive calls to left and right children

Summary

Visits left subtree, then right subtree, then the node itself.

Use when you want to process children before their parent, like deleting a tree or evaluating expressions.

The key is to print the node only after both children are fully processed.