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589. N 叉树的前序遍历

English Version

题目描述

给定一个 N 叉树，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）。

 

进阶：

递归法很简单，你可以使用迭代法完成此题吗?

 

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：

• N 叉树的高度小于或等于 1000
• 节点总数在范围 [0, 10^4] 内

解法

Python3

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""


class Solution:
    def preorder(self, root: 'Node') -> List[int]:
        ans = []
        if root is None:
            return ans
        stk = [root]
        while stk:
            node = stk.pop()
            ans.append(node.val)
            for child in node.children[::-1]:
                stk.append(child)
        return ans

Java

/*
// 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> preorder(Node root) {
        if (root == null) {
            return Collections.emptyList();
        }
        List<Integer> ans = new ArrayList<>();
        Deque<Node> stk = new ArrayDeque<>();
        stk.push(root);
        while (!stk.isEmpty()) {
            Node node = stk.pop();
            ans.add(node.val);
            List<Node> children = node.children;
            for (int i = children.size() - 1; i >= 0; --i) {
                stk.push(children.get(i));
            }
        }
        return ans;
    }
}

C++

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

    Node() {}

    Node(int _val) {
        val = _val;
    }

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

class Solution {
public:
    vector<int> preorder(Node* root) {
        if (!root) return {};
        vector<int> ans;
        stack<Node*> stk;
        stk.push(root);
        while (!stk.empty())
        {
            Node* node = stk.top();
            ans.push_back(node->val);
            stk.pop();
            auto children = node->children;
            for (int i = children.size() - 1; i >= 0; --i) stk.push(children[i]);
        }
        return ans;
    }
};

Go

/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Children []*Node
 * }
 */

func preorder(root *Node) []int {
	var ans []int
	if root == nil {
		return ans
	}
	stk := []*Node{root}
	for len(stk) > 0 {
		node := stk[len(stk)-1]
		ans = append(ans, node.Val)
		stk = stk[:len(stk)-1]
		children := node.Children
		for i := len(children) - 1; i >= 0; i-- {
			stk = append(stk, children[i])
		}
	}
	return ans
}

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