Leetcode - 590 - N-ary Tree Postorder Traversal (easy)

最新推荐文章于 2021-06-10 19:37:33 发布

原创最新推荐文章于 2021-06-10 19:37:33 发布 · 205 阅读

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本文提供了N叉树后序遍历的递归及迭代两种解决方案，包括Java、Python和C++等语言的实现代码，帮助读者理解并掌握不同编程语言下N叉树后序遍历的具体实现。

Recursive:

Java:

class Solution {
    List<Integer> list = new ArrayList<>();
    public List<Integer> postorder(Node root) {
        if (root == null)
            return list;
        
        for(Node node: root.children)
            postorder(node);
        
        list.add(root.val);
        
        return list;
    }
}

// https://leetcode.com/problems/n-ary-tree-postorder-traversal/discuss/147959/Java-Iterative-and-Recursive-Solutions

Python:

def postorder(self, root):
        """
        :type root: Node
        :rtype: List[int]
        """
        res = []
        if root == None: return res

        def recursion(root, res):
            for child in root.children:
                recursion(child, res)
            res.append(root.val)

        recursion(root, res)
        return res

# https://leetcode.com/problems/n-ary-tree-postorder-traversal/discuss/155806/Python-iterative-and-recursive-solution-with-explanation

Iterative:

add as preorder and then reverse

Java:

class Solution {
    public List<Integer> postorder(Node root) {
        List<Integer> list = new ArrayList<>();
        if (root == null) return list;
        
        Stack<Node> stack = new Stack<>();
        stack.add(root);
        
        while(!stack.isEmpty()) {
            root = stack.pop();
            list.add(root.val);
            for(Node node: root.children)
                stack.add(node);
        }
        Collections.reverse(list);
        return list;
    }
}

// https://leetcode.com/problems/n-ary-tree-postorder-traversal/discuss/147959/Java-Iterative-and-Recursive-Solutions

C++:

vector<int> postorder(Node* root) {
    if(root==NULL) return {};
    vector<int> res;
    stack<Node*> stk;
    stk.push(root);
    while(!stk.empty())
    {
        Node* temp=stk.top();
        stk.pop();
        for(int i=0;i<temp->children.size();i++) stk.push(temp->children[i]);
        res.push_back(temp->val);
    }
    reverse(res.begin(), res.end());
    return res;
}

// https://leetcode.com/problems/n-ary-tree-postorder-traversal/discuss/150339/C%2B%2B-super-easy-11-line-solution-beats-100!

Python:

list.extend() to add another list to the current one

def postorder(self, root):
        res = []
        if root == None: return res

        stack = [root]
        while stack:
            curr = stack.pop()
            res.append(curr.val)
            stack.extend(curr.children)

        return res[::-1]

# https://leetcode.com/problems/n-ary-tree-postorder-traversal/discuss/155806/Python-iterative-and-recursive-solution-with-explanation