Binary Tree Preorder Traversal
144. Binary Tree Preorder Traversal
Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?
class Solution {
func preorderTraversal(_ root: TreeNode?) -> [Int] {
var result: [Int] = []
helper(root, &result)
return result
}
func helper(_ root: TreeNode?, _ result: inout [Int]) {
guard let root = root else {
return
}
result.append(root.val)
helper(root.left, &result)
helper(root.right, &result)
}
}
/**
* Definition for a binary tree node.
* public class TreeNode {
* public var val: Int
* public var left: TreeNode?
* public var right: TreeNode?
* public init(_ val: Int) {
* self.val = val
* self.left = nil
* self.right = nil
* }
* }
*/
class Solution {
func preorderTraversal(_ root: TreeNode?) -> [Int] {
guard let root = root else {
return []
}
var result: [Int] = []
var stack: [TreeNode] = []
stack.append(root)
while !stack.isEmpty {
let node = stack.removeLast()
result.append(node.val)
if let right = node.right {
stack.append(right)
}
if let left = node.left {
stack.append(left)
}
}
return result
}
}
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<>();
if (root == null) {
return result;
}
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
result.add(node.val);
if (node.right != null) {
stack.push(node.right);
}
if (node.left != null) {
stack.push(node.left);
}
}
return result;
}
}
Hope this helps,
Michael