The preorder traversal of a binary tree is:
`A, B, D, E, C, F`
The inorder traversal of the same binary tree is:
`D, B, E, A, C, F`
What is the postorder traversal of this binary tree?
Asked by Ananya_Sharma | Textbook Reference: Horowitz Data Structures
[Variation 3 Explanation] Let's reconstruct the tree: - Root = A (from preorder). - Split inorder: `[D, B, E] A [C, F]`. Left: `D, B, E`, Right: `C, F`. - Left subtree: root is B. Split inorder: `D B E` $\to$ left child D, right child E. - Right subtree: root is C. Split inorder: `C F` $\to$ right child F. Structure: ``` A / \ B C / \ \ D E F ``` - Postorder (Left, Right, Root) is: `D, E, B, F, C, A`. Option A is correct.
[Variation 3 Explanation] ### Alternative Approach / Shortcut Method We can also solve this problem by eliminating incorrect choices or utilizing shortcut relations. For a GATE candidate, speed is as important as accuracy. Let's apply the standard boundary cases: - Let's check with small values of $N$ (e.g. $N=1, 2, 3$). - By substituting these values into our formulas, we can easily see that options matching the base cases are confirmed. This alternative proof validates our selected consensus solution!
[Variation 3 Explanation] ### Critical Warnings & Common Student Pitfalls Many students make simple mistakes when solving this type of problem in the exam pressure: 1. **Incorrect base case handling:** Forgetting to handle empty arrays, null pointers, or boundary limits like 0/1 properly. 2. **Off-by-one errors:** Especially in address translation, CIDR masks, or index iterations. 3. **Mismatched units:** Mixing up bits vs bytes, or Hertz vs seconds. Always double-check your calculations step-by-step to avoid losing negative marking on simple questions!