This was asked in LinkedIn Interview Given a list of child->parent relationships, build a binary tree out of it. All the element Ids inside the tree are unique. Example: Given the following relationships: Child Parent IsLeft 15 20 true 19 80 true 17 20 false 16 80 false 80 50 false 50 null false 20 50 true You should return the following tree: 50 / 20 80 / \ / 15 17 19 16 Function Signature /** * Represents a pair relation between one parent node and one child node inside a binary tree * If the _parent is null, it represents the ROOT node */ class Relation { int parent; int child; bool isLeft; }; /** * Represents a single Node inside a binary tree */ class Node { int id; Node * left; Node * right; } /** * Implement a method to build a tree from a list of parent-child relationships * And return the root Node of the tree */ Node * buildTree (vector<Relation> & data) { }
public class GenerateBinaryTree { public static class Relation { public Integer parent, child; public boolean isLeft; public Relation(Integer parent, Integer child, boolean isLeft) { this.parent = parent; this.child = child; this.isLeft = isLeft; } } public static class Node { public Integer val; public Node left, right; public Node(Integer val, Node left, Node right) { this.val = val; this.left = left; this.right = right; } } private static HashMap<Integer, Node> mapOfNodes; public static Node buildTree(List<Relation> relations) { mapOfNodes = new HashMap<Integer, Node>(); Iterator<Relation> relationsIterator = relations.iterator(); Relation relation; Node root = null; while (relationsIterator.hasNext()) { Node parentNode, childNode; relation = relationsIterator.next(); if (relation.parent == null) { root = getChildNode(relation.child); continue; } if (mapOfNodes.containsKey((relation.parent))) { parentNode = mapOfNodes.get(relation.parent); childNode = getChildNode(relation.child); if (relation.isLeft) parentNode.left = childNode; else parentNode.right = childNode; } else { childNode = getChildNode(relation.child); parentNode = new Node(relation.parent, relation.isLeft ? childNode : null, relation.isLeft ? null : childNode); mapOfNodes.put(relation.parent, parentNode); } } return root; } private static Node getChildNode(Integer child) { Node childNode; if (mapOfNodes.containsKey((child))) { childNode = mapOfNodes.get(child); } else { childNode = new Node(child, null, null); mapOfNodes.put(child, childNode); } return childNode; } }
时间: 2024-10-09 11:42:33