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429. N 叉树的层序遍历

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429. N 叉树的层序遍历

题目链接:429. N 叉树的层序遍历

题目描述

给定一个 N 叉树,返回其节点值的层序遍历。(即从左到右,逐层遍历)。

树的序列化输入是用层序遍历,每组子节点都由 null 值分隔(参见示例)。

示例 1:

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

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],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]

提示:

题解

思路:这道题与“二叉树”的层次遍历一样,只是孩子节点变多了而已。

代码(c++):

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<vector<int>> levelOrder(Node* root) {
        queue<Node*> que;
        if (root != nullptr) que.push(root);
        vector<vector<int>> result;
​
        while (!que.empty()) {
            int size = que.size();
            vector<int> temp;
            for (int i = 0; i < size; i++) {
                Node* node = que.front();
                que.pop();
                temp.push_back(node->val);
                for (int j = 0; j < node->children.size(); j++) {
                    if (node->children[j] != nullptr) que.push(node->children[j]);
                }
            }
            result.push_back(temp);
        }
        return result;
    }
};

代码(Java):

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 levelOrderSolution {
    public List<List<Integer>> levelOrder(Node root) {
        Deque<Node> que = new LinkedList<>();
        if (root != null) que.offer(root);
        List<List<Integer>> result = new ArrayList<>();
​
        while (!que.isEmpty()) {
            int size = que.size();
            List<Integer> temp = new ArrayList<>();
​
            for (int i = 0; i < size; i++) {
                Node node = que.poll();
                temp.add(node.val);
                for (Node child : node.children) {
                    if (child != null) que.offer(child);
                }
            }
​
            result.add(temp);
        }
        return result;
    }
}

分析:

标签:Node,遍历,val,int,层序,429,que,null,children
来源: https://www.cnblogs.com/wltree/p/15612133.html