notes/Leetcode Binary-Tree-Preorder-Traversal.md

Leetcode Binary-Tree-Preorder-Traversal

2022-07-04 14:51

Algorithms:

#algorithm #DFS #DFS_preorder

Data structures:

#DS #binary_tree

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

tag:#

Links:

• Link to problem

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Problem

Given the root of a binary tree, return the preorder traversal of its nodes' values.

Examples

Example 1:

Input: root = [1,null,2,3] Output: [1,2,3]

Example 2:

Input: root = [] Output: []

Example 3:

Input: root = [1] Output: [1]

Constraints

• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

Thoughts

[!summary] This is a #binary_tree #DFS_preorder problem.

Preorder, means root is at the "Pre" position, so the order is:

• Search root node
• Search left sub-tree
• Search right sub-tree The recursion version is easy to implement.

Solution

Recursion:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
  vector<int> preorderTraversal(TreeNode *root) {
    // If the node is empty
    if (root == nullptr) {
      return {};
    }
    // First check root
    vector<int> answer;

    answer.push_back(root->val);

    vector<int> left = preorderTraversal(root->left);
    answer.insert(answer.end(), left.begin(), left.end());

    vector<int> right = preorderTraversal(root->right);
    answer.insert(answer.end(), right.begin(), right.end());

    return answer;
  }
};