[LeetCode] Binary Tree Inorder Traversal

Binary Tree Inorder Traversal 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.
問題解決の考え方:
中順遍歴、再帰アルゴリズムに問題はありません.まず左サブツリー、それからルートノード、それから右サブツリーです.

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

反復バージョン:

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