binary-tree/traversal/binary-tree-inorder-traversal
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2023-12-01
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?
分析
用栈或者 Morris 遍历。
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;
}
}
// Binary Tree Inorder Traversal
// 使用栈，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
vector<int> inorderTraversal(TreeNode *root) {
vector<int> result;
stack<const TreeNode *> s;
const TreeNode *p = root;
while (!s.empty() || p != nullptr) {
if (p != nullptr) {
s.push(p);
p = p->left;
} else {
p = s.top();
s.pop();
result.push_back(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;
}
};