pre-order-traversal

Problem Description

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].
• -1000 <= Node.val <= 1000

Solution for Binary Tree Pre Order Traversal

Intuition

• Recursive is very easy try with iterative way

Code in Different Languages

Python

Written by @parikhitkurmi

//python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
 def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
     L=[]
     def pot(node):
         if node is None:
             return
         L.append(node.val)
         pot(node.left)
         pot(node.right)
         return L
     return pot(root)

Java

Written by @parikhitkurmi

//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 {
 List<Integer> result = new ArrayList<>();
 public List<Integer> preorderTraversal(TreeNode root) {
     preOrder(root);
     return result;
 }

 private void preOrder(TreeNode node){
     if(node == null){
         return;
     }

     result.add(node.val);
     preOrder(node.left);
     preOrder(node.right);
 }
}

C++

Written by @parikhitkurmi

//cpp

/**
* 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>ans ;
void Preorder(TreeNode* root) {
 if(!root) 
 return ;

 // NLR
 ans.push_back(root->val) ;
 Preorder(root->left);
 Preorder(root->right) ;

} 
 vector<int> preorderTraversal(TreeNode* root) {

     Preorder(root) ;
     return ans ;
     
 }
};

References

• LeetCode Problem: Binary Tree Preorder Traversal
• Solution Link: Binary Tree Preorder Traversal
• Authors GeeksforGeeks Profile: parikhit kurmi
• Authors Leetcode: parikhit kurmi