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
// 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;
}
};
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 preorderTraversal(TreeNode *root) {
vector 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;
}
};