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590. N-ary Tree Postorder Traversal

Description

Given the root of an n-ary tree, return the postorder 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: [5,6,3,2,4,1]

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: [2,6,14,11,7,3,12,8,4,13,9,10,5,1]

 

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

  • /*
    // 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> postorder(Node root) {
            LinkedList<Integer> ans = new LinkedList<>();
            if (root == null) {
                return ans;
            }
            Deque<Node> stk = new ArrayDeque<>();
            stk.offer(root);
            while (!stk.isEmpty()) {
                root = stk.pollLast();
                ans.addFirst(root.val);
                for (Node child : root.children) {
                    stk.offer(child);
                }
            }
            return ans;
        }
    }
    
  • /*
    // 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> postorder(Node* root) {
            vector<int> ans;
            if (!root) return ans;
            stack<Node*> stk{ {root} };
            while (!stk.empty()) {
                root = stk.top();
                ans.push_back(root->val);
                stk.pop();
                for (Node* child : root->children) stk.push(child);
            }
            reverse(ans.begin(), ans.end());
            return ans;
        }
    };
    
  • """
    # Definition for a Node.
    class Node:
        def __init__(self, val=None, children=None):
            self.val = val
            self.children = children
    """
    
    
    class Solution:
        def postorder(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:
                    stk.append(child)
            return ans[::-1]
    
    
  • /**
     * Definition for a Node.
     * type Node struct {
     *     Val int
     *     Children []*Node
     * }
     */
    
    func postorder(root *Node) []int {
    	var ans []int
    	if root == nil {
    		return ans
    	}
    	stk := []*Node{root}
    	for len(stk) > 0 {
    		root = stk[len(stk)-1]
    		stk = stk[:len(stk)-1]
    		ans = append([]int{root.Val}, ans...)
    		for _, child := range root.Children {
    			stk = append(stk, child)
    		}
    	}
    	return ans
    }
    
  • /**
     * Definition for node.
     * class Node {
     *     val: number
     *     children: Node[]
     *     constructor(val?: number) {
     *         this.val = (val===undefined ? 0 : val)
     *         this.children = []
     *     }
     * }
     */
    
    function postorder(root: Node | null): number[] {
        const res = [];
        const dfs = (root: Node | null) => {
            if (root == null) {
                return;
            }
            for (const node of root.children) {
                dfs(node);
            }
            res.push(root.val);
        };
        dfs(root);
        return res;
    }
    
    

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