Postorder historia
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This order makes deserialization (reconstruction) natural — the first element is the root, then comes its left and right trees in order.
Use Cases:
- Cloning or copying trees
- Serializing and deserializing trees
- Traversing a file system (e.g., printing folder names before their contents)
- Evaluating or printing expressions in prefix notation
- Assigning unique IDs top-down
Mental Model:
Here’s the leader.
Any traversal that lists every node in the tree exactly once is called an enumeration of the tree’s nodes. Now there role of it.
Problems
- Leet 105 Construct Binary Tree from Preorder and Inorder Traversal
- Leet 106 Construct Binary Tree from Inorder and Postorder Traversal
Så länge det funnits transporter har man kunnat beställa varor från andra orter, men först mot 1800-talets slut uppkom företag som helt inriktade sig på postorderförsäljning.
What’s rarely emphasized is this:
Table of Contents
A Quick Refresher
Given a binary tree, the three primary depth-first traversals are:
| Traversal | Order |
|---|---|
| Preorder | Root → Left → Right |
| Inorder | Left → Root → Right |
| Postorder | Left → Right → Root |
Preorder Traversal
I give you control first.
Key Idea: Root Comes First
Preorder is the only traversal that visits the root before anything else.
Here is an ADT for binary tree nodes, called . This is called a postorder traversal.
Example 12.3.2
The postorder enumeration for the tree of Figure 12.3.1 is D B G E H I F C A.
12.3.1.3. But to do that requires that the children’s children be deleted first, and so on. We should stop it when we got required values
Logic
Validate BST
Method 1: Inorder Traversal (Sorted Check)
- Key idea: If you do an inorder traversal of a valid BST, you get a strictly increasing sequence.
- Why it works:
- Inorder visits: Left → Root → Right
- which, in a BST, naturally produces values in ascending order.
- Time complexity: O(n)
- Space complexity: O(h) (stack space, h = tree height)
Method 2
- Time complexity: O(n)
- Space complexity: O(h)
Common Mistake to Avoid : Only checking immediate children is wrong.
At first glance:
But 6 is in the right subtree of 10, so it must be > 10 — rule is broken.
Range-check method catches this; naive child check won’t.
Build a Tree from traversal
For building back a tree from traversal we need
- preorder + inorder
- postorder + inorder
Refer
Inorder
Finding index(mid) of root of inorder will give
- Left subtree size = mid
- Right subtree size = len(inorder) – mid – 1
The mid from inorder.index(root) always gives you the size of the left subtree.
Case 1 — Preorder
- Root is always the first element (preorder[0]).
- The left subtree starts immediately after the root at index 1.
- Since left subtree size = mid, it takes mid elements:
Left sub tree
- starting index = 1
- end index = starting index+size = 1+m
Right sub tree
- strarting from 1+m, till last
Case 2 — Postorder
- Root is always the last element (postorder[-1]).
- The left subtree starts at index 0 in postorder and takes mid elements:
No matter preorder or postorder:
- mid = number of nodes in the left subtree
- In Preorder, left subtree starts at index 1.
- In Postorder, left subtree starts at index 0.
Example 1: Preorder + Inorder
Steps to solve
- Root = 3 (first in preorder)
- In inorder, 3 is at index 1 → left subtree size = 1
- Left preorder = [9]
- Right preorder = [20, 15, 7]
Example 2 — Postorder + Inorder
Steps to solve
- Root = 3 (last in postorder)
- In inorder, 3 is at index 1 → left size = 1
- Left postorder = [9]
- Right postorder = [15, 7, 20]
Examples
Code: Preorder + Inorder
Code: Postorder + Inorder
Where inorder(mid)?
we use inorder[:mid]) and inorder[mid+1:], leaving behind inorder[mid], as we already used that root.
For example, this would be necessary if we wish to return all nodes in the tree to free store. The key difference between them lies in the order in which the nodes are visited.
| Traversal Technique | Order of Visitation |
|---|---|
| Inorder Traversal | Left subtree -> Root -> Right |
| Preorder Traversal | Root -> Left subtree -> Right |
| Postorder Traversal | Left subtree -> Right subtree -> Root |
Implementing Postorder Traversal
In this section, we will delve into the step-by-step explanation of the postorder traversal algorithm and provide code examples in popular programming languages.
Step-by-Step Explanation of the Postorder Traversal Algorithm
The postorder traversal algorithm can be broken down into the following steps:
- Traverse the left subtree.
- Traverse the right subtree.
- Visit the root node.
For the recursive implementation, these steps are performed using recursive function calls.
So technically, all of these can clone. When you reach a leaf node, you finalize the path and store or return it.
Logic
- Traverse the tree
- Keep a running list of the path so far
- On reaching a leaf, record the current path
- Backtrack to explore other paths
- When you return from a child, you must remove its contribution to the current path.
- Without path.pop() or re-creating the path, the path would “leak” and be incorrect.
Implementation
Common problem
Why Use Preorder Traversal?
Preorder fits naturally here because:
- You need to process the current node first (add it to the path),
- Then recursively explore its children,
- And only at leaf nodes, decide to save the full path.
- This makes it a top-down traversal pattern.
Kth Smallest Element in BST
Inorder natually gives values in sorted form if it BST
If it’s just a binary tree (no BST property), inorder won’t necessarily be sorted.
So, we can use inorder, but we should’nt traversel entrire tree.
Ja de ligger fortfarande kvar i Finspång och säljer visitkort och stämplar.
Författare: Edward Blom
12.3. Some applications do not require that the nodes be visited in any particular order as long as each node is visited precisely once. It avoids the use of recursive function calls, making it more memory-efficient for large trees.Comparison with Other Tree Traversal Techniques
There are three primary tree traversal techniques: inorder, preorder, and postorder traversal.
It’s naturally fits
Prefix Expression (Polish Notation)
If you build an expression tree like:
PythonThe preorder traversal gives you the prefix expression: + * 3 4 5
Root-to-Leaf Path Tracking
In problems where you need to track paths from root to leaves, you usually start from the root and build the path as you recurse.
As you traverse the tree top-down (usually using preorder traversal), you maintain a path list or string that represents the nodes you’ve visited so far.
The order of visitation is: left subtree -> right subtree -> root node.

Postordern erbjöd inte bara nyttoföremål, utan all sorts lyx: tavlor, accessoarer, fotoalbum, skämtartiklar, prydnadsföremål etc. It tells you how left and right subtrees are arranged around each root.
Use Cases:
- Getting sorted data from BSTs
- Checking BST validity
- Printing expressions in human-readable infix format
- Splitting tree structure around a pivot
- Supporting reconstruction from preorder/inorder or postorder/inorder
Mental Model:
Let me understand the context on both sides before I act.
Postorder Traversal
I finish everything before you touch me.
Key Idea: Root Comes Last
Postorder defers visiting the root until both its children have been processed.
Den första varan, ett foto av kungafamiljen, kompletterades snart med hundratals andra.
Related Posts
Let me understand the context on both sides before I act.
I finish everything before you touch me.