590. N 叉树的后序遍历

590. N 叉树的后序遍历

1.递归

很easy

/*
// 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* rt) {
        vector<int>ans;
        function<void(Node*rt)> dfs = [&](Node*rt){
            if(rt==nullptr) return;
            for(auto son:rt->children){
                dfs(son);
            }
            ans.push_back(rt->val);  
        };
        dfs(rt);
        return ans;
    }
};

2.迭代

用栈，然后反后序，也就是右左根顺序，同时用一个标记数组维护该结点及子树是否都遍历过。

class Solution {
public:
    vector<int> postorder(Node* rt) {
        vector<int> res;
        if (rt == nullptr) {
            return res;
        }
        stack<Node*>st;
        unordered_map<Node*,bool>vis;
        st.push(rt);
        vector<int>b;
        while(!st.empty()){
            Node* u = st.top();
            if((int)u->children.size()==0 || vis[u]){
                b.push_back(u->val);
                st.pop();
                continue;
            }
            for(auto it = u->children.rbegin();it!=u->children.rend();it++){
                st.push(*it);
            }
            vis[u] = 1;
        }
        return b;
    }
};