186 lines
4.1 KiB
Markdown
186 lines
4.1 KiB
Markdown
# Leetcode Binary-Tree-Preorder-Traversal
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#### 2022-07-04 14:51
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> ##### Algorithms:
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> #algorithm #DFS #DFS_preorder #Morris_traversal
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> ##### Data structures:
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> #DS #binary_tree
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> ##### Difficulty:
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> #coding_problem #difficulty-easy
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> ##### Additional tags:
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> #leetcode #CS_list_need_understanding
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> ##### Revisions:
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> N/A
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##### Related topics:
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```expander
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tag:#DFS
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```
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- [[Leetcode Binary-Tree-Inorder-Traversal]]
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- [[Leetcode Binary-Tree-Postorder-Traversal]]
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- [[Leetcode Maximum-Depth-Of-Binary-Tree]]
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- [[Leetcode Symmetric-Tree]]
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##### Links:
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- [Link to problem](https://leetcode.com/problems/binary-tree-preorder-traversal/)
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___
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### Problem
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Given the `root` of a binary tree, return _the preorder traversal of its nodes' values_.
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#### Examples
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**Example 1:**
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**Input:** root = [1,null,2,3]
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**Output:** [1,2,3]
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**Example 2:**
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**Input:** root = []
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**Output:** []
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**Example 3:**
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**Input:** root = [1]
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**Output:** [1]
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#### Constraints
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- The number of nodes in the tree is in the range `[0, 100]`.
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- `-100 <= Node.val <= 100`
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### Thoughts
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> [!summary]
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> This is a #binary_tree #DFS_preorder problem.
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Preorder, means root is at the "Pre" position, so the order is:
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- Search root node
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- Search left sub-tree
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- Search right sub-tree
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The recursion version is easy to implement.
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Use Stacks for the iteration method, because stacks are FILO, the **subtree will get traversed first.**
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And remember to push **right subtree** first, so that the left one will be traversed first.
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The Morris traversal needs more understandings.
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### Solution
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Iteration
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
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* right(right) {}
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* };
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*/
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class Solution {
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public:
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vector<int> preorderTraversal(TreeNode *root) {
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if (!root) {
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return {};
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}
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// Using stacks, since FILO, the latest gets served first.
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vector<int> answer;
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stack<TreeNode *> pending;
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do {
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if (root != nullptr) {
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answer.push_back(root->val);
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// nothing gets pushed if is nullptr;
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pending.push(root->right);
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pending.push(root->left);
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}
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root = pending.top();
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pending.pop();
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} while (root || !pending.empty());
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return answer;
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}
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};
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```
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Recursion, using private funcs:
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
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* right(right) {}
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* };
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*/
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class Solution {
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public:
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vector<int> preorderTraversal(TreeNode *root) {
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// Use two functions:
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vector<int> answer;
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preorder(root, answer);
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return answer;
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}
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private:
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void preorder(TreeNode *root, vector<int> &answer) {
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if (root == nullptr) {
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return;
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}
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answer.push_back(root->val);
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preorder(root->left, answer);
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preorder(root->right, answer);
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}
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};
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```
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Recursion:
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
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* right(right) {}
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* };
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*/
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class Solution {
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public:
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vector<int> preorderTraversal(TreeNode *root) {
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// If the node is empty
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if (root == nullptr) {
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return {};
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}
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// First check root
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vector<int> answer;
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answer.push_back(root->val);
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vector<int> left = preorderTraversal(root->left);
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answer.insert(answer.end(), left.begin(), left.end());
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vector<int> right = preorderTraversal(root->right);
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answer.insert(answer.end(), right.begin(), right.end());
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return answer;
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}
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};
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```
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