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

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