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