Description#
Given the root
of an n-ary tree, return the preorder traversal of its nodes' values.
Nary-Tree input serialization is represented in their level order traversal. Each group of children is separated by the null value (See examples)
Example 1:
![](https://spcdn.pages.dev/leetcode/problems/0589.N-ary%20Tree%20Preorder%20Traversal/images/narytreeexample.png)
Input: root = [1,null,3,2,4,null,5,6]
Output: [1,3,5,6,2,4]
Example 2:
![](https://spcdn.pages.dev/leetcode/problems/0589.N-ary%20Tree%20Preorder%20Traversal/images/sample_4_964.png)
Input: 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]
Output: [1,2,3,6,7,11,14,4,8,12,5,9,13,10]
Constraints:
- The number of nodes in the tree is in the range
[0, 104]
. 0 <= Node.val <= 104
- The height of the n-ary tree is less than or equal to
1000
.
Follow up: Recursive solution is trivial, could you do it iteratively?
Solutions#
Solution 1#
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| """
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""
class Solution:
def preorder(self, root: 'Node') -> List[int]:
ans = []
if root is None:
return ans
stk = [root]
while stk:
node = stk.pop()
ans.append(node.val)
for child in node.children[::-1]:
stk.append(child)
return ans
|
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| /*
// 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<Integer> preorder(Node root) {
if (root == null) {
return Collections.emptyList();
}
List<Integer> ans = new ArrayList<>();
Deque<Node> stk = new ArrayDeque<>();
stk.push(root);
while (!stk.isEmpty()) {
Node node = stk.pop();
ans.add(node.val);
List<Node> children = node.children;
for (int i = children.size() - 1; i >= 0; --i) {
stk.push(children.get(i));
}
}
return ans;
}
}
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| /*
// 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<int> preorder(Node* root) {
if (!root) return {};
vector<int> ans;
stack<Node*> stk;
stk.push(root);
while (!stk.empty()) {
Node* node = stk.top();
ans.push_back(node->val);
stk.pop();
auto children = node->children;
for (int i = children.size() - 1; i >= 0; --i) stk.push(children[i]);
}
return ans;
}
};
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| /**
* Definition for a Node.
* type Node struct {
* Val int
* Children []*Node
* }
*/
func preorder(root *Node) []int {
var ans []int
if root == nil {
return ans
}
stk := []*Node{root}
for len(stk) > 0 {
node := stk[len(stk)-1]
ans = append(ans, node.Val)
stk = stk[:len(stk)-1]
children := node.Children
for i := len(children) - 1; i >= 0; i-- {
stk = append(stk, children[i])
}
}
return ans
}
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| /**
* Definition for node.
* class Node {
* val: number
* children: Node[]
* constructor(val?: number) {
* this.val = (val===undefined ? 0 : val)
* this.children = []
* }
* }
*/
function preorder(root: Node | null): number[] {
const res = [];
if (root == null) {
return res;
}
const stack = [root];
while (stack.length !== 0) {
const { val, children } = stack.pop();
res.push(val);
const n = children.length;
for (let i = n - 1; i >= 0; i--) {
stack.push(children[i]);
}
}
return res;
}
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| /**
* Definition for a Node.
* struct Node {
* int val;
* int numChildren;
* struct Node** children;
* };
*/
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
void dfs(struct Node* root, int* ans, int* i) {
if (!root) {
return;
}
ans[(*i)++] = root->val;
for (int j = 0; j < root->numChildren; j++) {
dfs(root->children[j], ans, i);
}
}
int* preorder(struct Node* root, int* returnSize) {
int* ans = malloc(sizeof(int) * 10000);
*returnSize = 0;
dfs(root, ans, returnSize);
return ans;
}
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Solution 2#
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| /**
* Definition for node.
* class Node {
* val: number
* children: Node[]
* constructor(val?: number) {
* this.val = (val===undefined ? 0 : val)
* this.children = []
* }
* }
*/
function preorder(root: Node | null): number[] {
const ans = [];
const dfs = (root: Node | null) => {
if (root == null) {
return;
}
ans.push(root.val);
for (const node of root.children) {
dfs(node);
}
};
dfs(root);
return ans;
}
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