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# Binary Tree Inorder Traversal LeetCode Solution

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 = 
Output: ``````

Constraints:

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

### Binary Tree Inorder Traversal LeetCode Solution in Java

``````public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();

Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode cur = root;

while(cur!=null || !stack.empty()){
while(cur!=null){
cur = cur.left;
}
cur = stack.pop();
cur = cur.right;
}

return list;
}
``````

### Binary Tree Inorder Traversal LeetCode Solution in Python

``````def preorder(root):
return [root.val] + preorder(root.left) + preorder(root.right) if root else []
``````

### Binary Tree Inorder Traversal LeetCode Solution in C++

``````class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> nodes;
stack<TreeNode*> todo;
while (root || !todo.empty()) {
while (root) {
todo.push(root);
root = root -> left;
}
root = todo.top();
todo.pop();
nodes.push_back(root -> val);
root = root -> right;
}
return nodes;
}
};
``````
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