Binary Tree Level Order Traversal

102. Binary Tree Level Order Traversal

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],

3
/ \
9  20
/  \
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