in-order-traversal

Problem Description

Given the root of a binary tree, return the inorder 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 inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:

     def InOrder(root,arr):
         
         if root is None:
             return

         else:
             InOrder(root.left,arr)
             arr.append(root.val)
             InOrder(root.right,arr)

         return arr

     return InOrder(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 {
 public List<Integer> inorderTraversal(TreeNode root) {
     
     List<Integer> ans = new ArrayList<>();
     ArrayDeque<TreeNode> stack = new ArrayDeque<>();

     TreeNode cur = root;

     while(cur != null || !stack.isEmpty()) {

         while(cur != null) {

             stack.push(cur);
             cur = cur.left;
         }

         TreeNode pop = stack.pop();
         ans.add(pop.val);

         cur = pop.right;
     }
     return ans;
 }
}

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 inorder (TreeNode* root) {

 // inorder(LNR)

 if(!root) 
 return ;



 inorder(root->left) ;


 ans.push_back(root->val);

 
 inorder(root->right) ;


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

     inorder(root) ;

      return ans ; 
     
 }
};

References

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