The “Reconstruct Itinerary” problem asks you to rebuild a travel route using a list of flight tickets. Each ticket represents a directed edge from one airport to another, and the itinerary must use every ticket exactly once. The route must begin at a specified starting airport, and when multiple valid itineraries exist, you must return the one that is lexicographically smallest when read as a sequence of airport codes.
This problem looks like a straightforward path-building exercise, but it quickly becomes tricky because of two constraints that fight each other. You must use every ticket exactly once, which creates a strict edge-usage requirement, and you must also satisfy a lexicographical ordering rule, which influences which edge you take when multiple choices are available.
Why greedy “always pick the smallest next airport” can fail
A natural instinct is to take the smallest available next destination at each step to satisfy the lexicographical requirement. Unfortunately, that can lead to dead ends where you can no longer use all remaining tickets. Even if a choice looks correct locally, it may strand you in a part of the graph that prevents you from traversing all edges.
This is an important signal: the problem is not simply about picking the next best-looking option. It is about finding a path that uses every edge exactly once. That requirement is the hallmark of an Eulerian path problem in a directed graph.
Recognizing the Eulerian path structure
When a problem asks you to traverse every edge exactly once, you should immediately think about Eulerian paths. Here, airports are nodes and tickets are directed edges. The itinerary is essentially an Eulerian path starting from the required starting node.
This reframing is the turning point. Instead of treating the task like general backtracking over permutations, you treat it as a structured graph traversal problem where there is a well-known constructive approach to guarantee that every edge is used exactly once.
Want to explore more coding problem solutions? Check out the Minimum Cost for Tickets and the Furthest Building You Can Reach.
Why Hierholzer’s algorithm is the right tool
The classic way to build an Eulerian path efficiently is a method commonly known as Hierholzer’s algorithm. The core idea is to keep walking edges until you cannot continue, then backtrack and stitch partial routes together. This backtracking is not the same as brute-force backtracking over choices. It is systematic and guarantees that edges are consumed in a way that produces a valid Eulerian path.
What makes this approach so effective is that it does not guess. It consumes edges, and when it reaches a dead end, it commits that node to the final itinerary order and returns to a previous branching point.
How lexicographical order fits into the traversal
The itinerary must be lexicographically smallest among all valid itineraries. That requirement affects the order in which you choose outgoing edges from each airport. To satisfy it, you always want to use the smallest available destination first when leaving a node.
The key nuance is that “choose smallest” must happen within a traversal that is already correct for Eulerian paths. Hierholzer-style traversal provides correctness, and lexicographical ordering is enforced by ensuring each airport’s outgoing destinations are processed in sorted order. This way, whenever there is a choice, the traversal consumes the smallest edge first, but still preserves the ability to use all edges because the underlying algorithm handles dead ends properly.
Why the route is built in reverse
One of the most surprising aspects of this problem is that the itinerary is typically constructed in reverse order during traversal. When you reach a node with no remaining outgoing edges, that node is appended to the route. This is effectively “postorder” with respect to edges: you add airports to the final answer only when you are done using all tickets that depart from them.
This reversal is exactly what allows partial cycles and branches to be merged into one valid route. Once traversal finishes, reversing the constructed list yields the correct forward itinerary.
Why this approach guarantees correctness
Correctness comes from two guarantees. First, the Eulerian-path construction ensures every ticket is used exactly once because edges are consumed and never reused. Second, the lexicographical requirement is satisfied because at each airport, edges are selected in sorted order, ensuring that among all valid Eulerian itineraries, the one produced is the smallest in lexical sequence.
Importantly, this avoids the exponential blow-up of naive backtracking. You are not exploring all possible itineraries; you are deterministically constructing the correct one.
Time and space complexity considerations
The runtime is dominated by organizing outgoing flights in sorted order and then traversing each ticket exactly once. Space is used to store the adjacency structure for the graph and the route being built.
This efficiency is what makes the approach practical. Even when there are many tickets, the algorithm remains scalable because it does not branch exponentially.
Why this problem is popular in interviews
Reconstruct Itinerary is a classic interview problem because it tests whether candidates can recognize a hidden Eulerian path structure and combine it with lexicographical constraints. Many candidates get stuck trying brute-force permutations or naive greedy choices, both of which can fail.
The problem rewards a deeper understanding of graph traversal and the ability to apply the right algorithmic model rather than forcing a generic search.
What this problem teaches beyond flight routes
Beyond the travel story, this problem teaches a valuable lesson about edge-based traversal problems. When you must use every edge exactly once, you are often in Eulerian territory, and the solution usually involves structured traversal with controlled backtracking rather than brute force.
If you can clearly explain why greedy alone fails, how the Eulerian-path perspective unlocks a constructive solution, and how lexicographical order is enforced without sacrificing correctness, you demonstrate strong algorithmic maturity. That depth of understanding makes “Reconstruct Itinerary” an excellent exercise in graph modeling and deterministic path construction.
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