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429. N 叉树的层序遍历

English Version

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：

  • 树的高度不会超过 1000
  • 树的节点总数在 [0, 10^4] 之间

解法

“BFS 层次遍历实现”。

Python3

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


class Solution:
    def levelOrder(self, root: 'Node') -> List[List[int]]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            t = []
            for _ in range(len(q), 0, -1):
                root = q.popleft()
                t.append(root.val)
                q.extend(root.children)
            ans.append(t)
        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<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Deque<Node> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            List<Integer> t = new ArrayList<>();
            for (int n = q.size(); n > 0; --n) {
                root = q.poll();
                t.add(root.val);
                q.addAll(root.children);
            }
            ans.add(t);
        }
        return ans;
    }
}

也可以使用 DFS:

class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> res = new ArrayList<>();
        dfs(root, 0, res);
        return res;
    }

    private void dfs(Node root, int level, List<List<Integer>> res) {
        if (root == null) {
            return;
        }
        if (res.size() <= level) {
            res.add(new ArrayList<>());
        }
        res.get(level++).add(root.val);
        for (Node child : root.children) {
            dfs(child, level, res);
        }
    }
}

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<vector<int>> levelOrder(Node* root) {
        vector<vector<int>> ans;
        if (!root) return ans;
        queue<Node*> q{{root}};
        while (!q.empty())
        {
            vector<int> t;
            for (int n = q.size(); n > 0; --n)
            {
                root = q.front();
                q.pop();
                t.push_back(root->val);
                for (auto& child : root->children) q.push(child);
            }
            ans.push_back(t);
        }
        return ans;
    }
};

Go

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

func levelOrder(root *Node) [][]int {
	var ans [][]int
	if root == nil {
		return ans
	}
	q := []*Node{root}
	for len(q) > 0 {
		var t []int
		for n := len(q); n > 0; n-- {
			root = q[0]
			q = q[1:]
			t = append(t, root.Val)
			for _, child := range root.Children {
				q = append(q, child)
			}
		}
		ans = append(ans, t)
	}
	return ans
}

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