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?
分析
用栈或者 Morris 遍历。
栈
- Python
- Java
- C++
// 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;
}
}
// Binary Tree Preorder Traversal
// 使用栈，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {
vector<int> result;
stack<const TreeNode *> s;
if (root != nullptr) s.push(root);
while (!s.empty()) {
const TreeNode *p = s.top();
s.pop();
result.push_back(p->val);
if (p->right != nullptr) s.push(p->right);
if (p->left != nullptr) s.push(p->left);
}
return result;
}
};
# Binary Tree Preorder Traversal
# Using stack, time complexity O(n), space complexity O(n)
class Solution:
def preorderTraversal(self, root):
result = []
s = []
if root:
s.append(root)
while s:
p = s.pop()
result.append(p.val)
if p.right:
s.append(p.right)
if p.left:
s.append(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;
}
};