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Formatted question description: https://leetcode.ca/all/590.html

590. N-ary Tree Postorder Traversal (Easy)

Given an n-ary tree, return the postorder traversal of its nodes' values.

For example, given a 3-ary tree:

 

 

Return its postorder traversal as: [5,6,3,2,4,1].

 

Note:

Recursive solution is trivial, could you do it iteratively?

Related Topics:
Tree

Similar Questions:

Solution 1. Recursive

// OJ: https://leetcode.com/problems/n-ary-tree-postorder-traversal/
// Time: O(N)
// Space: O(logN)
class Solution {
private:
    vector<int> ans;
    void rec(Node *root) {
        if (!root) return;
        for (auto ch : root->children) rec(ch);
        ans.push_back(root->val);
    }
public:
    vector<int> postorder(Node* root) {
        rec(root);
        return ans;
    }
};

Solution 2. Iterative

Preorder traverse the mirrored tree and reverse the result!

// OJ: https://leetcode.com/problems/n-ary-tree-postorder-traversal/
// Time: O(N)
// Space: O(logN)
class Solution {
public:
    vector<int> postorder(Node* root) {
        if (!root) return {};
        stack<Node*> s;
        s.push(root);
        vector<int> ans;
        while (s.size()) {
            root = s.top();
            s.pop();
            ans.push_back(root->val);
            for (auto c : root->children) s.push(c);
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};
  • /*
    // 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) {
            List<Integer> postorder = new ArrayList<Integer>();
            Stack<Node> stack1 = new Stack<Node>();
            Stack<Node> stack2 = new Stack<Node>();
            if (root != null)
                stack1.push(root);
            while (!stack1.isEmpty()) {
                Node temp = stack1.pop();
                stack2.push(temp);
                List<Node> children = temp.children;
                int size = children.size();
                for (int i = 0; i < size; i++)
                    stack1.add(children.get(i));
            }
            while (!stack2.isEmpty())
                postorder.add(stack2.pop().val);
            return postorder;
        }
    }
    
    ############
    
    /*
    // 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;
        }
    }
    
  • // OJ: https://leetcode.com/problems/n-ary-tree-postorder-traversal/
    // Time: O(N)
    // Space: O(logN)
    class Solution {
    private:
        vector<int> ans;
        void rec(Node *root) {
            if (!root) return;
            for (auto ch : root->children) rec(ch);
            ans.push_back(root->val);
        }
    public:
        vector<int> postorder(Node* root) {
            rec(root);
            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.
    class Node(object):
        def __init__(self, val, children):
            self.val = val
            self.children = children
    """
    class Solution(object):
        def postorder(self, root):
            """
            :type root: Node
            :rtype: List[int]
            """
            res = []
            if not root:
                return res
            for child in root.children:
                res.extend(self.postorder(child))
            res.append(root.val)
            return res
    
  • /**
     * 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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