LeetCode 589. N 叉树的前序遍历

589. N 叉树的前序遍历

给定一个 n 叉树的根节点  root ，返回 其节点值的 前序遍历 。

n 叉树 在输入中按层序遍历进行序列化表示，每组子节点由空值 null 分隔（请参见示例）。

示例 1：

输入：root = [1,null,3,2,4,null,5,6]
输出：[1,3,5,6,2,4]

示例 2：

输入：root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
输出：[1,2,3,6,7,11,14,4,8,12,5,9,13,10]

提示：

• 节点总数在范围 [0, 10^4]内
• 0 <= Node.val <= 10^4
• n 叉树的高度小于或等于 1000

进阶：递归法很简单，你可以使用迭代法完成此题吗?

解法思路：

1、递归（Recursion）

2、迭代（Iterator）

法一：

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

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List preorder(Node root) {
        // Recursion
        // Time: O(n) n 为节点数
        // Space: O(n)
        List res = new ArrayList<>();
        helper(root, res);
        return res;
    }

    private void helper(Node root, List res) {
        if (root == null) return;
        res.add(root.val);
        for (Node node : root.children) {
            helper(node, res);
        }
    }
}

法二：

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

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List preorder(Node root) {
        // Iterator
        // Time: O(n) n 为节点数
        // Space: O(n)
        List res = new ArrayList();
        if (root == null)
            return res;
        Map map = new HashMap();
        Deque stack = new ArrayDeque();
        Node node = root;
        while (!stack.isEmpty() || node != null) {
            while (node != null) {
                res.add(node.val);
                stack.push(node);
                List children = node.children;
                if (children != null && children.size() > 0) {
                    map.put(node, 0);
                    node = children.get(0);
                } else {
                    node = null;
                }
            }
            node = stack.peek();
            int index = map.getOrDefault(node, -1) + 1;
            List children = node.children;
            if (children != null && children.size() > index) {
                map.put(node, index);
                node = children.get(index);
            } else {
                stack.pop();
                map.remove(node);
                node = null;
            }
        }
        return res;
    }
}

迭代优化：

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

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List preorder(Node root) {
        // Optimize Iterator
        // Time: O(n) n 为节点数
        // Space: O(n)
        List res = new ArrayList<>();
        if (root == null) {
            return res;
        }

        Deque stack = new ArrayDeque();
        stack.push(root);
        while (!stack.isEmpty()) {
            Node node = stack.pop();
            res.add(node.val);
            for (int i = node.children.size() - 1; i >= 0; --i) {
                stack.push(node.children.get(i));
            }
        }
        return res;
    }
}