LeetCode #1719 — HARD

Number Of Ways To Reconstruct A Tree

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given an array pairs, where pairs[i] = [xi, yi], and:

  • There are no duplicates.
  • xi < yi

Let ways be the number of rooted trees that satisfy the following conditions:

  • The tree consists of nodes whose values appeared in pairs.
  • A pair [xi, yi] exists in pairs if and only if xi is an ancestor of yi or yi is an ancestor of xi.
  • Note: the tree does not have to be a binary tree.

Two ways are considered to be different if there is at least one node that has different parents in both ways.

Return:

  • 0 if ways == 0
  • 1 if ways == 1
  • 2 if ways > 1

A rooted tree is a tree that has a single root node, and all edges are oriented to be outgoing from the root.

An ancestor of a node is any node on the path from the root to that node (excluding the node itself). The root has no ancestors.

Example 1:

Input: pairs = [[1,2],[2,3]]
Output: 1
Explanation: There is exactly one valid rooted tree, which is shown in the above figure.

Example 2:

Input: pairs = [[1,2],[2,3],[1,3]]
Output: 2
Explanation: There are multiple valid rooted trees. Three of them are shown in the above figures.

Example 3:

Input: pairs = [[1,2],[2,3],[2,4],[1,5]]
Output: 0
Explanation: There are no valid rooted trees.

Constraints:

  • 1 <= pairs.length <= 105
  • 1 <= xi < yi <= 500
  • The elements in pairs are unique.
Patterns Used
🌳Tree Traversal→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an array pairs, where pairs[i] = [xi, yi], and: There are no duplicates. xi < yi Let ways be the number of rooted trees that satisfy the following conditions: The tree consists of nodes whose values appeared in pairs. A pair [xi, yi] exists in pairs if and only if xi is an ancestor of yi or yi is an ancestor of xi. Note: the tree does not have to be a binary tree. Two ways are considered to be different if there is at least one node that has different parents in both ways. Return: 0 if ways == 0 1 if ways == 1 2 if ways > 1 A rooted tree is a tree that has a single root node, and all edges are oriented to be outgoing from the root. An ancestor of a node is any node on the path from the root to that node (excluding the node itself). The root has no ancestors.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Tree

Example 1

[[1,2],[2,3]]

Example 2

[[1,2],[2,3],[1,3]]

Example 3

[[1,2],[2,3],[2,4],[1,5]]

Related Problems

  • Create Binary Tree From Descriptions (create-binary-tree-from-descriptions)
  • Maximum Star Sum of a Graph (maximum-star-sum-of-a-graph)
Step 02

Core Insight

What unlocks the optimal approach

  • Think inductively. The first step is to get the root. Obviously, the root should be in pairs with all the nodes. If there isn't exactly one such node, then there are 0 ways.
  • The number of pairs involving a node must be less than or equal to that number of its parent.
  • Actually, if it's equal, then there is not exactly 1 way, because they can be swapped.
  • Recursively, given a set of nodes, get the node with the most pairs, then this must be a root and have no parents in the current set of nodes.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1719: Number Of Ways To Reconstruct A Tree
class Solution {
    public int checkWays(int[][] pairs) {
        boolean[][] g = new boolean[510][510];
        List<Integer>[] v = new List[510];
        Arrays.setAll(v, k -> new ArrayList<>());
        for (int[] p : pairs) {
            int x = p[0], y = p[1];
            g[x][y] = true;
            g[y][x] = true;
            v[x].add(y);
            v[y].add(x);
        }
        List<Integer> nodes = new ArrayList<>();
        for (int i = 0; i < 510; ++i) {
            if (!v[i].isEmpty()) {
                nodes.add(i);
                g[i][i] = true;
            }
        }
        nodes.sort(Comparator.comparingInt(a -> v[a].size()));
        boolean equal = false;
        int root = 0;
        for (int i = 0; i < nodes.size(); ++i) {
            int x = nodes.get(i);
            int j = i + 1;
            for (; j < nodes.size() && !g[x][nodes.get(j)]; ++j)
                ;
            if (j < nodes.size()) {
                int y = nodes.get(j);
                if (v[x].size() == v[y].size()) {
                    equal = true;
                }
                for (int z : v[x]) {
                    if (!g[y][z]) {
                        return 0;
                    }
                }
            } else {
                ++root;
            }
        }
        if (root > 1) {
            return 0;
        }
        return equal ? 2 : 1;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(h)

Approach Breakdown

LEVEL ORDER
O(n) time
O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL
O(n) time
O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.

Related Problems
#105Construct Binary Tree from Preorder and Inorder Traversalmedium→#106Construct Binary Tree from Inorder and Postorder Traversalmedium→#690Employee Importancemedium→#889Construct Binary Tree from Preorder and Postorder Traversalmedium→#1110Delete Nodes And Return Forestmedium→