Leetcode 429. N-ary Tree Level Order Traversal

最新推荐文章于 2025-01-12 10:14:09 发布

SnailTyan

最新推荐文章于 2025-01-12 10:14:09 发布

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分类专栏： Leetcode 文章标签： Leetcode

本文链接：https://blog.youkuaiyun.com/Quincuntial/article/details/81516906

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本文由Tyan撰写，详细介绍了N-ary树的层次遍历算法，包括递归和迭代两种实现方式，提供了清晰的C++代码示例，并引用了LeetCode上的题目链接供读者进一步了解。

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文章作者：Tyan
博客：noahsnail.com | 优快云 | 简书

1. Description

N-ary Tree Level Order Traversal

2. Solution

Recursive

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

    Node() {}

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        vector<vector<int>> result;
        if(!root) {
            return result;
        }
        queue<Node*> level;
        level.push(root);
        levelOrderTraverse(result, level);
        return result;
    }

    void levelOrderTraverse(vector<vector<int>>& result, queue<Node*> level) {
        if(level.empty()) {
            return;
        }
        queue<Node*> nextLevel;
        vector<int> values;
        while(!level.empty()) {
            Node* current = level.front();
            level.pop();
            values.push_back(current->val);
            for(int i = 0; i < current->children.size(); i++) {
                nextLevel.push(current->children[i]);
            }
        }
        result.push_back(values);
        levelOrderTraverse(result, nextLevel);
    }
};

Iterative

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

    Node() {}

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        vector<vector<int>> result;
        if(!root) {
            return result;
        }
        queue<Node*> level;
        level.push(root);
        levelOrderTraverse(result, level);
        return result;
    }

    void levelOrderTraverse(vector<vector<int>>& result, queue<Node*> level) {
        if(level.empty()) {
            return;
        }
        while(!level.empty()) {
            queue<Node*> nextLevel;
            vector<int> values;
            while(!level.empty()) {
                Node* current = level.front();
                level.pop();
                values.push_back(current->val);
                for(int i = 0; i < current->children.size(); i++) {
                    nextLevel.push(current->children[i]);
                }
            }
            result.push_back(values);
            level = nextLevel;
        }
    }

};