# Binary Tree Preorder Traversal

### 描述​

Given a binary tree, return the preorder traversal of its nodes' values.

For example: Given binary tree {1,#,2,3},

 1  \   2  / 3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

### 栈​

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

### Morris 先序遍历​

// Binary Tree Preorder Traversal// Morris先序遍历，时间复杂度O(n)，空间复杂度O(1)class Solution {public:    vector<int> preorderTraversal(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刚刚被访问过 */                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 */                    result.push_back(cur->val); /* 仅这一行的位置与中序不同 */                    node->right = cur;                    prev = cur; /* cur刚刚被访问过 */                    cur = cur->left;                } else {    /* 已经线索化，则删除线索  */                    node->right = nullptr;                    /* prev = cur; 不能有这句，cur已经被访问 */                    cur = cur->right;                }            }        }        return result;    }};