# Binary Tree Inorder Traversal

### 描述​

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?

### 栈​

// Binary Tree Inorder Traversal// 使用栈，时间复杂度O(n)，空间复杂度O(n)class Solution {    public List<Integer> inorderTraversal(TreeNode root) {        ArrayList<Integer> result = new ArrayList<>();        Stack<TreeNode> s = new Stack<>();        TreeNode p = root;        while (!s.empty() || p != null) {            if (p != null) {                s.push(p);                p = p.left;            } else {                p = s.pop();                result.add(p.val);                p = p.right;            }        }        return result;    }}

### Morris 中序遍历​

// Binary Tree Inorder Traversal// Morris中序遍历，时间复杂度O(n)，空间复杂度O(1)class Solution {public:    vector<int> inorderTraversal(TreeNode *root) {        vector<int> result;        TreeNode *cur = root, *prev = nullptr;        while (cur != nullptr) {            if (cur->left == nullptr) {                result.push_back(cur->val);                prev = cur;                cur = cur->right;            } else {                /* 查找前驱 */                TreeNode *node = cur->left;                while (node->right != nullptr && node->right != cur)                    node = node->right;                if (node->right == nullptr) { /* 还没线索化，则建立线索 */                    node->right = cur;                    /* prev = cur; 不能有这句，cur还没有被访问 */                    cur = cur->left;                } else {    /* 已经线索化，则访问节点，并删除线索  */                    result.push_back(cur->val);                    node->right = nullptr;                    prev = cur;                    cur = cur->right;                }            }        }        return result;    }};