(LeetCode) 102. Binary Tree Level Order Traversal

102. Binary Tree Level Order Traversal

• Explore : Interview > Top Interveiw Questions > Easy Collection
• 분류 : Trees
• 난이도 : Medium

Problem

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

Example 1

Input: root = [1,2,2,3,4,4,3]
Output: true

Example 2

Input: root = [1,2,2,null,3,null,3]
Output: false

Constraints

• The number of nodes in the tree is in the range [1, 1000].
• -100 <= Node.val <= 100

Solution

Exapnder

/**
 * Example:
 * var ti = TreeNode(5)
 * var v = ti.`val`
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
class Solution {
    fun levelOrder(root: TreeNode?): List<List<Int>> {
        var result: MutableList<List<Int>> = mutableListOf()
        if (root == null) {
            return result
        }
        
        var queue: Queue<TreeNode> = LinkedList()
        queue.add(root)
        
        while (queue.isNotEmpty()) {
            var elementArray: MutableList<Int> = mutableListOf()
            for (i in 0 until queue.size) {
                var node: TreeNode = queue.poll()
                elementArray.add(node.`val`)
                node.left?.let {
                    queue.add(it)
                }
                node.right?.let {
                    queue.add(it)
                }
            }
            result.add(elementArray)
        }
        return result
    }
}

Point of Thinking

• Level 하나당 queue에 모든 노드를 쌓은 뒤, 한 번 순회할때마다 리스트를 만들어서 반환한다.