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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。

// 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;
}
};

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