LeetCode-N-ary Tree Preorder Traversal

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Description：
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?

题意：返回一颗N叉树的前序遍历节点值；

解法一（递归）：将根节点的值插入到list中，对根节点，我们依次遍历他的所有子节点，对每一个子节点递归调用自身；

Java

/*
// 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 {
    List<Integer> result = new ArrayList<>();
    public List<Integer> preorder(Node root) {
        if (root == null) return result;
        result.add(root.val);
        for (Node node : root.children) {
            preorder(node);
        }
        return result;
    }
}

解法二（非递归）：要想使用非递归实现前序遍历，我们需要利用一个栈来存储子节点；对每一个节点，我们从他的最后节点依次到最左节点存入到栈中；每一次，我们取出一个栈中元素，将其值插入到list中，再将其所有子节点从右往左依次保存到栈中；直到最后的栈为空；

Java

/*
// 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> result = new ArrayList<>();
        Stack<Node> treeNode = new Stack<>();
        if (root == null) return result;
        treeNode.push(root);
        while (!treeNode.isEmpty()) {
            Node temp = treeNode.pop();
            result.add(temp.val);
            for (int i = temp.children.size() - 1; i >= 0; i--) {
                treeNode.push(temp.children.get(i));
            }
        }
        return result;
    }
}