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
“BFS 层次遍历实现”。
"""
# 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
/*
// 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);
}
}
}