589. N-ary Tree Preorder Traversal

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:

Input: root = [1,null,3,2,4,null,5,6]
Output: [1,3,5,6,2,4]

Example 2:

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

Python Code

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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

Java Code

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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;
    }
}

C++ Code

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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;
    }
};

Go Code

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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
}

TypeScript Code

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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;
}

C Code

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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;
}

Solution 2

TypeScript Code

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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;
}

589. N-ary Tree Preorder Traversal#

Description#

Solutions#

Solution 1#

Solution 2#

589. N-ary Tree Preorder Traversal

Description

Solutions

Solution 1

Solution 2