使用BFS克隆图形（无向）

Question

I wrote a simple algorithm to clone an undirected graph using BFS, but there seem to be some logic wrong that I can't figure out. Can some one take a look?

The idea is to visit each node and copy them only once, after copying a node, check if its neighbor is not copied, enqueue that neighbor; if its neighbor is already copied, put them into each others' neighbor vector.

UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
    //key -> old node, value -> the new copy
    unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> m;
    //the queue always contains old nodes that haven't been copied
    queue<UndirectedGraphNode*> q;
    if(node)
        q.push(node);

    while(!q.empty()) {

        UndirectedGraphNode* n = q.front();
        q.pop();
        if(m.count(n)) continue; // if the node is already copied, continue

        // create the copy
        m[n] = new UndirectedGraphNode(n->label);

        // loop through the neighbors, if it's copied already, add the new copy to new copy's neighbor list
        for(UndirectedGraphNode* oldNei : n->neighbors) {

            if(m.count(oldNei)) {
                UndirectedGraphNode* newNei = m[oldNei];
                m[n]->neighbors.push_back(newNei);
                newNei->neighbors.push_back(m[n]);
            }

            else// if not in the map, it's not copied/visited yet

                q.push(oldNei);
        }

    }

    return m[node];
}

Here's the node struct:

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */

Here's the OJ I'm practicing with: https://leetcode.com/problems/clone-graph/

Answer 1

First of all you have to handle NULL node case separately because in your code you will return m[NULL] if input graph is NULL which can cause runtime error .

Have a look at below code I have commented where I have added and deleted the code and why I have added and deleted the code.

UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
    // ++ if node is null just return null.
    if(node == NULL) return NULL;

    unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> m;
    queue<UndirectedGraphNode*> q;
    // -- if(node) no need to check this as we have checked it above
    q.push(node);

    // ++ as first input will not have copy so no need to check it in map
    // just create a copy and associate it with original node in map.
    UndirectedGraphNode *graphCopy = new UndirectedGraphNode(node->label);
    m[node] = graphCopy;

    while(!q.empty()) {
        UndirectedGraphNode* n = q.front();
        q.pop();
        // -- if(m.count(n)) continue; :- no need to check this condition as we will
        // only push non copied node in queue.

        // -- m[n] = new UndirectedGraphNode(n->label); :- we will create it inside loop

        // loop through the neighbors, if it's copied already, add the new copy to new copy's neighbor list
        for(UndirectedGraphNode* oldNei : n->neighbors) {
            // if node is processed/ copied earlier then just push it in neighbour of current node
            if(m.count(oldNei)) {
                UndirectedGraphNode* newNei = m[oldNei];
                m[n]->neighbors.push_back(newNei);
                // -- newNei->neighbors.push_back(m[n]); no need of making back connection as 
                // this node has already processed and contains all it neighbour
            }

            else {// if not in the map, it's not copied/visited yet then create new copy of node here 
                  //and push it into queue
                UndirectedGraphNode *p = new UndirectedGraphNode(oldNei->label); // ++ make a copy of node
                m[n]->neighbors.push_back(p); // ++ push it to current node neighbour
                m[oldNei] = p; // ++ associate copied node with its original one
                q.push(oldNei); // push that node to queue
            }
        }
    }
    return graphCopy;
} 

Edit:-

---

Where your program is wrong:-

1. For self-cycle you will be getting Wrong output. As you make self-cycle twice by making back link to node.

2. It is not necessary that if node A is having neighbor B then B will also have node A as neighbor. But for every node you are just pushing it neighbor but also pushing node in neighbor's neighbor, which is wrong.

3. If Node are not processed then you are just pushing it into queue not making any link between current node and unprocessed node.

For more details see execution of your code :-

Let's take same graph as of problem:
First node is labeled as 1. Connect node 1 to both nodes 2 and 3.
Second node is labeled as 2. Connect node 2 to node 3.
Third node is labeled as 3. Connect node 3 to node 3 (itself), thus   forming a self-cycle.

So original graph will have neighbor like this:
Neighbor 1 -> 2 , 3
Neighbor 2 -> 3
Neighbor 3 -> 3

       1
      / \
     /   \
    2 --- 3
         / \
         \_/

Now starting executing your code:
assume 1 as root node

first it will insert 1 in Queue
Q -> [ 1 ]

inside While Loop :-
----------------------------------------------------------------------------------------------------
First Iteration:-
n = 1
Pop : Q -> empty // 1 is poped out so queue is empty now

here node 1 is not in map so you will create a copy of node 1 and save it in map
Map -> (1 , 1C)

Check for neighbour of 1 -> which is 2 & 3

inside for loop both node 2 & 3 is not processed so you will push it in Queue so Queue will became:-

Q -> [2, 3]
----------------------------------------------------------------------------------------------------

Second Iteration:-
n = 2
Pop : Q -> [ 3 ] 

Node 2 is not in map-
Create copy of node 2 and push it in map

Map -> (1, 1C) , (2, 2C)

Check for neighbour of 2 -> which is 3

node 3 is not processed till now so push it in Queue 

Q -> [3, 3]

----------------------------------------------------------------------------------------------------

Third Tteration:-

n = 3
Pop : Q -> [ 3 ]

node 3 is not in map so-
Create copy of node 3 and push it in map

Map -> (1, 1C) , (2, 2C) , (3, 3C)

Check for neighbour of 3 -> which is 3

Node 3 is also in map so you will create a link between node 3C & 3C which is correct but you will do it twice as you are pushing back node also so in this step you will do as follows

Neighbour 3C -> 3C
Neighbour 3C -> 3C

So over all it will look like

Neighbour 1C -> 
Neighbour 2C -> 
Neighbour 3C -> 3C, 3C

This is your final result and it different from original
----------------------------------------------------------------------------------------------------