94. 二叉树的中序遍历

给定一个二叉树的根节点 root ，返回 它的 中序 遍历 。

示例 1：

输入：root = [1,null,2,3]
输出：[1,3,2]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1]
输出：[1]

提示：

• 树中节点数目在范围 [0, 100] 内
• -100 <= Node.val <= 100

/**
 * 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 inorder(TreeNode* root, vector<int>& vec) {
        if (root == nullptr) {
            return;
        }
        inorder(root->left, vec);
        vec.push_back(root->val);
        inorder(root->right, vec);
    }
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> vec;
        inorder(root, vec);
        return vec;
    }
};

/**
 * 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> inorderTraversal(TreeNode* root) {
        stack<TreeNode*> stk;
        vector<int> vec;
        while (root != nullptr || !stk.empty()) {
	        while (root != nullptr) {
		        stk.push(root);
		        root = root->left;
	        }
	        root = stk.top();
	        stk.pop();
	        vec.push_back(root->val);
	        root = root->right;
        }
        return vec;
    }
};