Binary Tree Level Order Traversal

Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

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Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public IList<IList<int>> LevelOrder(TreeNode root) {
        
        IList<IList<int>> output = new List<IList<int>>();
        if(root!=null){
        Queue<TreeNode> queue = new Queue<TreeNode>();
        queue.Enqueue(root);
        while (queue.Count != 0)
        {
            int size= queue.Count;
            var ans=new List<int>();
            for (int i = 0; i < size; i++)
            {
                TreeNode node = queue.Dequeue();
                if (node.left!=null)
                    queue.Enqueue(node.left);
                if (node.right!=null)
                    queue.Enqueue(node.right);

                ans.Add(node.val);
            }
            output.Add(ans);
        }
        }
        return output;
    }
}

Let's analyze the time and space complexity of the given algorithm:

• Time Complexity: The time complexity of this algorithm is O(n), where n is the number of nodes in the binary tree. This is because the algorithm needs to visit all the nodes of the tree to perform the level order traversal.

• Space Complexity: The space complexity of this algorithm is O(n), where n is the number of nodes in the binary tree. This is because in the worst case, when the tree is a complete binary tree, the maximum number of nodes at any level is n/2, so the queue would need to store n/2 nodes at once. The space required for the result list is also proportional to n, so it doesn't affect the overall space complexity.