133. Clone Graph

Problem:

Given the head of a graph, return a deep copy (clone) of the graph. Each node in the graph contains a label (int) and a list (List[UndirectedGraphNode]) of its neighbors. There is an edge between the given node and each of the nodes in its neighbors.

OJ’s undirected graph serialization (so you can understand error output):

Nodes are labeled uniquely.

We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.

As an example, consider the serialized graph {0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

Visually, the graph looks like the following:

       1
      / \
     /   \
    0 --- 2
         / \
         \_/

Note: The information about the tree serialization is only meant so that you can understand error output if you get a wrong answer. You don’t need to understand the serialization to solve the problem.

Solution:

DFS. Cache the visited node before entering the next recursion.

/**
 * Definition for undirected graph.
 * function UndirectedGraphNode(label) {
 *     this.label = label;
 *     this.neighbors = [];   // Array of UndirectedGraphNode
 * }
 */

/**
 * @param {UndirectedGraphNode} graph
 * @return {UndirectedGraphNode}
 */
let cloneGraph = function (graph) {
  const cache = {};
  return _clone(graph);

  function _clone(graph) {
    if (!graph) {
      return graph;
    }
    const label = graph.label;
    if (!cache[label]) {
      cache[label] = new UndirectedGraphNode(label);
      cache[label].neighbors = graph.neighbors.map(_clone);
    }
    return cache[label];
  }
};

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