【java】589. N-ary Tree Preorder Traversal

问题原文如下https://leetcode-cn.com/problems/n-ary-tree-preorder-traversal/description/

递归法

public List<Integer> preorder(Node root) {
        List<Integer> res = new LinkedList<>();
        prehel(root,res);
        return res;
    }
	
	public void prehel(Node root,List<Integer> res) {
		if(root == null) return;
		List<Node> child = root.children;
		res.add(root.val);
		for(Node ele:child) prehel(ele,res);
	}

非递归法

//用于返回结果
        List<Integer> res = new LinkedList<>();    
        //用于回退
        Stack<Node> stack = new Stack<>();
        //判断节点是否遍历过，如果遍历过，指针就要往后挪动一位，并且以此来改变pos
        Map<Node,Integer> map = new HashMap<>();
        
        
        //对两种特殊情况先进行处理
        if(root == null) return res;
        if(root.children.size() == 0) {
        	res.add(root.val);
        	return res;
        }
        
        Node tmp = root;
        int pos = -1;
        
        while(!stack.isEmpty() || tmp != null) {
        	//判断当前节点的孩子是否存在，如果存在就要对孩子进行遍历
        	if(tmp.children.size()>0) {
        		if(!map.containsKey(tmp)) {
        			res.add(tmp.val);
        			map.put(tmp,0);
        			pos = 0;
        		}else {
        			map.put(tmp,map.get(tmp)+1);
        			pos = map.get(tmp);
        		}
        		stack.add(tmp);
        		if(pos<tmp.children.size())
        			tmp = tmp.children.get(pos);
        		else {
        			stack.pop();
        			if(!stack.isEmpty()) 
        				tmp = stack.pop();
        			else
        				break;
        		}
        	}else {
        		res.add(tmp.val);
        		tmp = stack.pop();
        	}
        }

        return res;