LeetCode

Binary Tree Level Order Traversal

Michael

Apr 22, 2017 • 1 min read

LintCode-69.Binary Tree Level Order Traversal

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

For example:
Given binary tree [3,9,20,null,null,15,7],

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

/**
 * 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 levelOrder(_ root: TreeNode?) -> [[Int]] {
        guard let root = root else {
            return []
        }
        
        var result: [[Int]] = []
        
        var queue: [TreeNode] = []
        queue.append(root)
        
        while !queue.isEmpty {
            let count = queue.count
            var list: [Int] = []
            
            for _ in 0..<count {
                let node = queue.removeFirst()
                list.append(node.val)
                
                if let left = node.left {
                    queue.append(left)
                }
                if let right = node.right {
                    queue.append(right)
                }
            }
            
            result.append(list)
        }
        
        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<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new ArrayList<>();
            
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                list.add(node.val);
                
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            
            result.add(list);
        }
        
        return result;
    }
}

Hope this helps,
Michael