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133. Clone Graph

Given a reference of a node in a connected undirected graph.

Return a deep copy (clone) of the graph.

Each node in the graph contains a value (int) and a list (List[Node]) of its neighbors.

class Node {
    public int val;
    public List<Node> neighbors;
}

Test case format:

For simplicity, each node’s value is the same as the node’s index (1-indexed). For example, the first node with val == 1, the second node with val == 2, and so on. The graph is represented in the test case using an adjacency list.

An adjacency list is a collection of unordered lists used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.

The given node will always be the first node with val = 1. You must return the copy of the given node as a reference to the cloned graph.

Example 1:

Input: adjList = [[2,4],[1,3],[2,4],[1,3]]
Output: [[2,4],[1,3],[2,4],[1,3]]
Explanation: There are 4 nodes in the graph.
1st node (val = 1)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
2nd node (val = 2)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).
3rd node (val = 3)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
4th node (val = 4)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).

Example 2:

Input: adjList = [[]]
Output: [[]]
Explanation: Note that the input contains one empty list. The graph consists of only one node with val = 1 and it does not have any neighbors.

Example 3:

Input: adjList = []
Output: []
Explanation: This an empty graph, it does not have any nodes.

Constraints:

Solution

Code Steps

"""
# Definition for a Node.
class Node:
    def __init__(self, val = 0, neighbors = None):
        self.val = val
        self.neighbors = neighbors if neighbors is not None else []
"""
class Solution:
    def cloneGraph(self, root: 'Node') -> 'Node':
        # corner case
        if root is None:
            return None
        # clone nodes by BFS
        old2new = dict() # key: old node, val: new node
        q = deque([root])
        qed = set(q)
        while q:
            # pop
            old_node = q.popleft()
            # Process
            old2new[old_node] = Node(val=old_node.val)
            # append children
            for old_nei in old_node.neighbors:
                if old_nei not in qed:
                    q.append(old_nei)
                    qed.add(old_nei)
        # clone edge
        for old_node in old2new:
            for old_nei in old_node.neighbors:
                old2new[old_node].neighbors.append(old2new[old_nei])
        return old2new[root]

要点

Advanced Solution

Use qed as record, no extra dixtionary

"""
# Definition for a Node.
class Node:
    def __init__(self, val = 0, neighbors = None):
        self.val = val
        self.neighbors = neighbors if neighbors is not None else []
"""
class Solution:
    def cloneGraph(self, root: 'Node') -> 'Node':
        # corner case
        if root is None:
            return None
        # clone nodes by BFS
        q = deque([root])
        qed = {root: Node(val=root.val)} # key: old node, val: new node
        while q:
            # pop
            old_node = q.popleft()
            # append children
            for old_nei in old_node.neighbors:
                if old_nei not in qed:
                    q.append(old_nei)
                    qed[old_nei] = Node(val=old_nei.val) # record enqueue and process
        # clone edge
        for old_node in qed:
            for old_nei in old_node.neighbors:
                qed[old_node].neighbors.append(qed[old_nei])
        return qed[root]