題目描述：

給定一個二叉樹，返回它的中序 遍歷。

示例:

輸入: [1,null,2,3]
   1
    \
     2
    /
   3

輸出: [1,3,2]

進階: 遞迴演算法很簡單，你可以通過迭代演算法完成嗎？

AC C++ Solution: 1.遞迴法：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

/* 遞迴法 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> nodes;
        inorder(root,nodes);
        return nodes;
    }
    
private:
    void inorder(TreeNode* root, vector<int>& nodes) {
        if(!root)
            return;
        
        inorder(root->left,nodes);  //當左子節點為空時,訪問當前節點值
        nodes.push_back(root->val);
        inorder(root->right,nodes);
    }
};
 

2.堆疊法：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

/* 堆疊法*/
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> vector;
        stack<TreeNode *> stack;
        TreeNode *pCurrent = root;
        
        while(!stack.empty() || pCurrent)
        {
           if(pCurrent) {
               stack.push(pCurrent);
               pCurrent =  pCurrent->left;
           }
            else {
                pCurrent = stack.top();
                stack.pop();
                vector.push_back(pCurrent->val);
                pCurrent = pCurrent->right;
            }
        }
        return vector;
    }
};
};