590. N 叉树的后序遍历

Posted 2022-03-15 Harris-H

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;
    
;