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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]]:
        if root is None:
            return []
        q = collections.deque([root])
        res = []
        while q:
            n = len(q)
            t = []
            for _ in range(n):
                node = q.popleft()
                t.append(node.val)
                if node.children:
                    q.extend(node.children)
            res.append(t)
        return res

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) {
        if (root == null) {
            return Collections.emptyList();
        }
        Deque<Node> q = new ArrayDeque<>();
        List<List<Integer>> res = new ArrayList<>();
        q.offer(root);
        while (!q.isEmpty()) {
            List<Integer> t = new ArrayList<>();
            for (int i = 0, n = q.size(); i < n; ++i) {
                Node node = q.poll();
                t.add(node.val);
                if (node.children != null) {
                    q.addAll(node.children);
                }
            }
            res.add(t);
        }
        return res;
    }
}

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);
        }
    }
}

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