LeetCode_94_二叉树的中序遍历

题目描述：

给定一个二叉树的根节点 root ，返回它的 中序 遍历。

示例：
输入：root = [1,null,2,3]
输出：[1,3,2]

输入：root = []
输出：[]

输入：root = [1]
输出：[1]

输入：root = [1,2]
输出：[2,1]

输入：root = [1,null,2]
输出：[1,2]

提示：
1. 树中节点数目在范围 [0, 100] 内
2. -100 <= Node.val <= 100

解法1：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:

    vector<int> res;  //容器要定义在外面

    vector<int> inorderTraversal(TreeNode *root) {

        if (root == nullptr) return {};

        inorderTraversal(root->left);
        res.push_back(root->val);
        inorderTraversal(root->right);

        return res;
    }
};

解法2：栈

/**
 * 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) {

        vector<int> ret;  //定义一个容器
        stack<TreeNode* > s;  //定义一个栈

        TreeNode* cur = root;

        while (cur != NULL || !s.empty()) {

            if (cur != NULL) { 

                s.push(cur); 
                cur = cur->left;           
            } else {

                cur = s.top(); 
                s.pop();
                ret.push_back(cur->val);  
                cur = cur->right;        
            }
        }

        return ret;
    }
};