力扣-145题二叉树的后序遍历（C++）- 递归+迭代

﹏ヽ暗。殇╰゛Y 2023-01-23 08:54 13阅读 0赞

题目链接：[https://leetcode-cn.com/problems/binary-tree-postorder-traversal/][https_leetcode-cn.com_problems_binary-tree-postorder-traversal]  
题目如下：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQwNDY3Njcw_size_16_color_FFFFFF_t_70]  
思路一：递归

/**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
     *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
     *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
     * };
     */
    class Solution {
        
    public:
        void postorder(vector<int>& result,TreeNode* root){
        //后序遍历：左右根
            if(root==NULL) return ;
            
            postorder(result,root->left);
            postorder(result,root->right);
            result.push_back(root->val);
        }
    
        vector<int> postorderTraversal(TreeNode* root) {
        
            vector<int> result;
            postorder(result,root);
    
            return result;
        }
    };

思路二：迭代

/**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
     *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
     *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
     * };
     */
    class Solution {
        
    public:
        vector<int> postorderTraversal(TreeNode* root) {
        
            //后序遍历顺序：左右根
            //在栈中，根左右，弹出顺序：右左根
            vector<int> result;
            stack<TreeNode*> stk;
    
            if(root==NULL) return result;
            stk.push(root);
    
            while(!stk.empty()){
        
                TreeNode* node=stk.top();//根
                stk.pop();
    
                result.push_back(node->val);
    
                if(node->left) stk.push(node->left);//左
                if(node->right) stk.push(node->right);//右
            }
    
            reverse(result.begin(),result.end());//result中的数据为根右左，reverse之后，变成左右根
    
            return result;
        }
    };

[https_leetcode-cn.com_problems_binary-tree-postorder-traversal]: https://leetcode-cn.com/problems/binary-tree-postorder-traversal/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQwNDY3Njcw_size_16_color_FFFFFF_t_70]: /images/20221004/00d524ac855442e485cdb415e7b2bc20.png