LeetCode 590. N 叉树的后序遍历

本文介绍: 590. N 叉树的后序遍历

590. N 叉树的后序遍历

给定一个 n 叉树的根节点 root ，返回其节点值的 后序遍历 。

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：[5,6,3,2,4,1]

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：[2,6,14,11,7,3,12,8,4,13,9,10,5,1]

提示：

节点总数在范围 [0, 104] 内
0 <= Node.val <= 104
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）效率低

法一：

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

    public Node() {}

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

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

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

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

法二：

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

    public Node() {}

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

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

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

优化迭代：

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

    public Node() {}

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

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

class Solution {
    public List<Integer> postorder(Node root) {
        // Optimize Iterator
        // Time: O(n), n 为节点数
        // Space: O(n)
        List<Integer> res = new ArrayList<>();
        if (root == null) return res;
        Deque<Node> stack = new ArrayDeque<>();
        Set<Node> visited = new HashSet<Node>();
        stack.addFirst(root);
        while (!stack.isEmpty()) {
            Node node = stack.peek();
            if (node.children.isEmpty() || visited.contains(node)) {
                stack.removeFirst();
                res.add(node.val);
                continue;
            }
            for (int i = node.children.size() - 1; i >= 0; --i) {
                stack.addFirst(node.children.get(i));
            }
            visited.add(node);
        }
        return res;
    }
}