Flip Binary Tree To Match Preorder Traversal

Flip Binary Tree To Match Preorder Traversal - Problem

You are given the root of a binary tree with n nodes, where each node is uniquely assigned a value from 1 to n. You are also given a sequence of n values voyage, which is the desired pre-order traversal of the binary tree.

Any node in the binary tree can be flipped by swapping its left and right subtrees. For example, flipping node 1 will swap its left and right children.

Flip the smallest number of nodes so that the pre-order traversal of the tree matches voyage. Return a list of the values of all flipped nodes. You may return the answer in any order.

If it is impossible to flip the nodes in the tree to make the pre-order traversal match voyage, return the list [-1].

Input & Output

Example 1 — Need to Flip Root

$ Input: root = [1,2], voyage = [1,2]

› Output: []

💡 Note: Tree already matches voyage [1,2] with preorder traversal, no flips needed

Example 2 — Flip Required

$ Input: root = [1,2,3], voyage = [1,3,2]

› Output: [1]

💡 Note: Original preorder is [1,2,3], but we need [1,3,2]. Flip node 1 to swap children 2 and 3

Example 3 — Impossible Case

$ Input: root = [1,2,3], voyage = [1,2,4]

› Output: [-1]

💡 Note: Tree has node 3 but voyage expects node 4, which doesn't exist in tree

Constraints

The number of nodes in the tree is n
n == voyage.length
1 ≤ n ≤ 100
1 ≤ Node.val, voyage[i] ≤ n
All the values in the tree are unique
All the values in voyage are unique

Visualization

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Asked in

G Google 15 a Amazon 8 M Microsoft 5

The key insight is to perform a preorder traversal while comparing with the expected voyage. When the left child doesn't match the expected next value, we need to flip the current node. The optimal approach uses DFS with early decision making: Time O(n), Space O(n).

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

DFS with Backtracking

⏱️ Time: O(2^n) Space: O(n)

Traverse the tree in preorder while comparing with the expected voyage. When we encounter a mismatch, try flipping the current node and continue. Backtrack if no solution is possible.

Try All Possible Flips

⏱️ Time: O(2^n × n) Space: O(n)

For each node, try both flipped and non-flipped states. Generate all 2^n combinations and check which ones produce the desired preorder traversal.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Smart Actions

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