DFS&Iteration Binary Tree Inorder Traversal

思路：
二叉树的中序遍历。

方法一：DFS。

/** * 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 {private:    void dfs(vector<int>& ans, TreeNode* node) {        if(node == nullptr) return;        if(node->left) {            dfs(ans, node->left);        }        ans.push_back(node->val);        if(node->right) {            dfs(ans, node->right);        }    }public:    vector<int> inorderTraversal(TreeNode* root) {        vector<int> ans;        dfs(ans, root);        return ans;    }};

方法二：迭代。
时间复杂度O(N)，空间复杂度O(N)

/** * 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> ans;        TreeNode* p = root;        stack<TreeNode*> s;        while(!s.empty() || p != nullptr) {            if(p != nullptr) {                s.push(p);                p = p->left;            }else {                p = s.top();                s.pop();                ans.push_back(p->val);                p = p->right;            }        }        return ans;    }};

方法三：