Clone Graph

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

BFS: 用一个HashMap存储已经生成过的点，避免重复生成相同的点。

public class Solution {
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if (node == null) {
            return null;
        }
        
        HashMap<UndirectedGraphNode, UndirectedGraphNode> map = 
        new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
        Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
        queue.offer(node);
        map.put(node, new UndirectedGraphNode(node.label));
        while (!queue.isEmpty()) {
            UndirectedGraphNode cur = queue.poll();
            UndirectedGraphNode newcur = map.get(cur);
            for (UndirectedGraphNode neig : cur.neighbors) {
                //if contains, this node has already been searched
                if (!map.containsKey(neig)) {
                    map.put(neig, new UndirectedGraphNode(neig.label));
                    queue.offer(neig);
                }
                newcur.neighbors.add(map.get(neig));
            }
        }
        return map.get(node);
    }
}

Binary Tree Level Order Traversal I && II

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

用一个queue进行BFS。第二个只需把25行改成add(0, level)

public class Solution {
    //Time: O(n)  Space: O(n)
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if (root == null) {
            return res;
        }
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            ArrayList<Integer> level = new ArrayList<Integer>();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                level.add(cur.val);
                
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            res.add(level);
        }
        return res;
    }
}