LeetCode589. N-ary Tree Preorder Traversal（N叉树的前序遍历）JAVA实现

Given an n-ary tree, return the preorder traversal of its nodes' values.

For example, given a 3-ary tree:

Return its preorder traversal as: [1,3,5,6,2,4].

Note:

Recursive solution is trivial, could you do it iteratively?

解题思路一：递归实现

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val,List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public List<Integer> preorder(Node root) {
        List<Integer> res = new ArrayList<Integer>();
        preOrder(res,root);
        return res;
    }
    public void preOrder(List<Integer> res, Node root){
        if(root==null) return;
        res.add(root.val);
        for(int i=0;i<root.children.size();i++)
            preOrder(res,root.children.get(i));
    }
}

解题思路二：非递归遍历。前序的非递归遍历容易实现，即先访问根，再将子结点从右往左压栈。

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val,List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public List<Integer> preorder(Node root) {
        List<Integer> res = new ArrayList<Integer>();
        Stack<Node> st = new Stack<Node>();
        st.push(root);
        Node p = null;
        while(st.empty()==false){
            p = st.pop();
            if(p!=null){
                res.add(p.val);
                for(int i=p.children.size()-1;i>=0;i--){
                    st.push(p.children.get(i));
                }
            }
        }
        return res;
    }
}