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# Binary Tree Preorder Traversal LeetCode Solution

## Problem – Binary Tree Preorder Traversal LeetCode Solution

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]``````

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 Preorder Traversal LeetCode Solution in Java

``````public List<Integer> preorderTraversal(TreeNode node) {
Stack<TreeNode> rights = new Stack<TreeNode>();
while(node != null) {
if (node.right != null) {
rights.push(node.right);
}
node = node.left;
if (node == null && !rights.isEmpty()) {
node = rights.pop();
}
}
return list;
}
``````

### Binary Tree Preorder Traversal LeetCode Solution in Python

``````def preorderTraversal(self, root):
ret = []
stack = [root]
while stack:
node = stack.pop()
if node:
ret.append(node.val)
stack.append(node.right)
stack.append(node.left)
return ret
``````

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

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