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Hello **Programmers/Coders,** Today we are going to share ** solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python**. At Each Problem with Successful submission with

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In this post, you will find the solution for the **Binary Tree Inorder Traversal** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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- Mathematics/Basic Logical Based Questions
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- Depth-First Search
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** Link for the Problem** – Binary Tree Inorder Traversal– LeetCode Problem

Binary Tree Inorder Traversal– LeetCode Problem

**Problem:**

Given the `root`

of a binary tree, return *the inorder traversal of its nodes’ values*.

**Example 1:**

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

**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`

Binary Tree Inorder Traversal– LeetCode Solutions

Binary Tree Inorder TraversalSolution in C++:

class Solution { public: vectorinorderTraversal(TreeNode* root) { vector ans; stack stack; while (root || !stack.empty()) { while (root) { stack.push(root); root = root->left; } root = stack.top(), stack.pop(); ans.push_back(root->val); root = root->right; } return ans; } };

Binary Tree Inorder Traversal Solution in Java:

class Solution { public ListinorderTraversal(TreeNode root) { List ans = new ArrayList<>(); Stack stack = new Stack<>(); while (root != null || !stack.isEmpty()) { while (root != null) { stack.push(root); root = root.left; } root = stack.pop(); ans.add(root.val); root = root.right; } return ans; } }

Binary Tree Inorder Traversal Solution in Python:

class Solution: def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]: ans = [] stack = [] while root or stack: while root: stack.append(root) root = root.left root = stack.pop() ans.append(root.val) root = root.right return ans