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
方法一:BFS
借助队列,逐层遍历。
时间复杂度
方法二:DFS
按深度遍历。
假设当前深度为 i,遍历到的节点为 root。若结果列表 ans[i]
不存在,则创建一个空列表放入 ans 中,然后将 root.val
放入 ans[i]
。接着往下一层遍历(root 的子节点)。
时间复杂度
BFS:
"""
# 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)):
root = q.popleft()
t.append(root.val)
q.extend(root.children)
ans.append(t)
return ans
DFS:
"""
# 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]]:
def dfs(root, i):
if root is None:
return
if len(ans) <= i:
ans.append([])
ans[i].append(root.val)
for child in root.children:
dfs(child, i + 1)
ans = []
dfs(root, 0)
return ans
BFS:
/*
// 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;
}
}
DFS:
/*
// 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<>();
dfs(root, 0, ans);
return ans;
}
private void dfs(Node root, int i, List<List<Integer>> ans) {
if (root == null) {
return;
}
if (ans.size() <= i) {
ans.add(new ArrayList<>());
}
ans.get(i++).add(root.val);
for (Node child : root.children) {
dfs(child, i, ans);
}
}
}
BFS:
/*
// 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;
}
};
DFS:
/*
// 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;
dfs(root, 0, ans);
return ans;
}
void dfs(Node* root, int i, vector<vector<int>>& ans) {
if (!root) return;
if (ans.size() <= i) ans.push_back({});
ans[i++].push_back(root->val);
for (Node* child : root->children) dfs(child, i, ans);
}
};
BFS:
/**
* 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
}
DFS:
/**
* Definition for a Node.
* type Node struct {
* Val int
* Children []*Node
* }
*/
func levelOrder(root *Node) [][]int {
var ans [][]int
var dfs func(root *Node, i int)
dfs = func(root *Node, i int) {
if root == nil {
return
}
if len(ans) <= i {
ans = append(ans, []int{})
}
ans[i] = append(ans[i], root.Val)
for _, child := range root.Children {
dfs(child, i+1)
}
}
dfs(root, 0)
return ans
}
BFS:
/**
* Definition for node.
* class Node {
* val: number
* children: Node[]
* constructor(val?: number) {
* this.val = (val===undefined ? 0 : val)
* this.children = []
* }
* }
*/
function levelOrder(root: Node | null): number[][] {
const res = [];
if (root == null) {
return res;
}
const queue = [root];
while (queue.length !== 0) {
const n = queue.length;
const vals = [];
for (let i = 0; i < n; i++) {
const { val, children } = queue.shift();
vals.push(val);
queue.push(...children);
}
res.push(vals);
}
return res;
}
DFS:
/**
* Definition for node.
* class Node {
* val: number
* children: Node[]
* constructor(val?: number) {
* this.val = (val===undefined ? 0 : val)
* this.children = []
* }
* }
*/
function levelOrder(root: Node | null): number[][] {
const res = [];
const dfs = (root: Node | null, depth: number) => {
if (root == null) {
return;
}
if (res.length <= depth) {
res.push([]);
}
const { val, children } = root;
res[depth].push(val);
children.forEach(node => dfs(node, depth + 1));
};
dfs(root, 0);
return res;
}