# Binary Tree Preorder Traversal – Leetcode Solution

In this post, we are going to solve the 144. Binary Tree Preorder Traversal problem of Leetcode. This problem 144. Binary Tree Preorder Traversal is a Leetcode easy level problem. Let’s see the code, 144. Binary Tree Preorder Traversal – Leetcode Solution.

## Problem

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

### Example 1 :

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

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

Now, let’s see the code of 144. Binary Tree Preorder Traversal – Leetcode Solution.

# Binary Tree Preorder Traversal – Leetcode Solution

### 144. Binary Tree Preorder Traversal – Solution in Java

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {

public void preOrder(TreeNode root, List<Integer> ans){
if(root == null) return;

preOrder(root.left,ans);
preOrder(root.right,ans);
}
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> ans = new ArrayList<Integer>();
preOrder(root,ans);
return ans;
}
}

### 144. Binary Tree Preorder Traversal – Solution in C++

/**
* 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) {
vector<int> v;
preTraversal(root, v);
return v;
}
void preTraversal(TreeNode* root, vector<int>& v){
if(!root) return;
v.push_back(root->val);
preTraversal(root->left, v);
preTraversal(root->right, v);
}
};

### 144. Binary Tree Preorder Traversal– Solution in Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
def preorderTraversal(self, root):
res = []
self.dfs(root, res)
return res

def dfs(self, root, res):
if root:
res.append(root.val)
self.dfs(root.left, res)
self.dfs(root.right, res)

Note: This problem 144. Binary Tree Preorder Traversal is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.