Binary Tree Preorder Traversal 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 = [1] Output: [1]
Constraints:
- The number of nodes in the tree is in the range
[0, 100]
. -100 <= Node.val <= 100
Binary Tree Preorder Traversal Solutions
✅Time: O(n)
✅Space: O(h)
C++
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> ans;
preorder(root, ans);
return ans;
}
private:
void preorder(TreeNode* root, vector<int>& ans) {
if (!root)
return;
ans.push_back(root->val);
preorder(root->left, ans);
preorder(root->right, ans);
}
};
Java
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> ans = new ArrayList<>();
preorder(root, ans);
return ans;
}
private void preorder(TreeNode root, List<Integer> ans) {
if (root == null)
return;
ans.add(root.val);
preorder(root.left, ans);
preorder(root.right, ans);
}
}
Python
class Solution:
def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
ans = []
def preorder(root: Optional[TreeNode]) -> None:
if not root:
return
ans.append(root.val)
preorder(root.left)
preorder(root.right)
preorder(root)
return ans
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