binary-tree-inorder-traversal 귀속과 비귀속 중 순서가 두 갈래 나무를 옮겨다니다

제목 설명

Given a binary tree, return the inorder traversal of its nodes' values.
For example: Given binary tree{1,#,2,3},

   1
    \
     2
    /
   3

return[1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".
귀속

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector ret;
    vector inorderTraversal(TreeNode *root) {
        helper(root);
        return ret;
    }
    
    void helper(TreeNode* root){
        if(!root) return;
        helper(root->left);
        ret.push_back(root->val);
        helper(root->right);
    }
};

비귀속

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector inorderTraversal(TreeNode *root) {
        vector ret;
        stack s;
        TreeNode* cur = root;
        if(!root) return ret;
        while(!s.empty() || cur){
            if(cur) {
                s.push(cur);
                cur = cur->left;
            }
            else {
                cur = s.top(); s.pop();
                ret.push_back(cur->val);
                cur = cur->right;
            }
        }
        return ret;
    }
};