LeetCode #590. N-ary Tree Postorder Traversal

题目描述：

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].

class Solution {
public:
    vector<int> postorder(Node* root) {
        if(root==NULL) return {};
        if(root->children.size()==0) return {root->val};
        vector<int> result;
        unordered_map<Node*,int> next_child;
        stack<Node*> s;
        s.push(root);
        Node* p=root;
        while(!s.empty())
        {
            if(next_child.count(p)==0) next_child[p]=0;
            int i=next_child[p];
            next_child[p]++;
            if(i==p->children.size()) 
            {
                result.push_back(p->val);
                next_child.erase(p);
                p=s.top();
                s.pop();
            }
            else
            {
                s.push(p);
                p=p->children[i];
            }
        }
        return result;
    }
};

class Solution {
public:
    vector<int> postorder(Node* root) {
        if(root==NULL) return {};
        vector<int> result;
        stack<Node*> s;
        s.push(root);
        // 后序遍历先访问第一个子节点到最后一个子节点，然后是父节点
        // 用栈模拟的时候可以先访问父节点，再访问最后一个子节点，到第一个子节点
        while(!s.empty())
        {
            Node* cur=s.top();
            s.pop();
            result.push_back(cur->val);
            for(int i=0;i<cur->children.size();i++)
                s.push(cur->children[i]); // 正着将所有子节点压栈，那么栈顶是最后一个子节点
        }
        // 最后将结果翻转得到正确的顺序
        reverse(result.begin(),result.end());
        return result;
    }
};