[LeetCode] 102. Binary Tree Level Order Traversal（二叉树的中序遍历）

• Difficulty: Medium

• Related Topics: Tree, Breadth-first Search

• Link: https://leetcode.com/problems/binary-tree-level-order-traversal/

Description

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
给定一棵二叉树，返回其层序遍历（从左至右，从上至下）。

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]
]

Solution

/**
 * 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
 * }
 */
import java.util.*

class Solution {
    fun levelOrder(root: TreeNode?): List<List<Int>> {
        if (root == null) {
            return emptyList()
        }
        val queue: Queue<TreeNode> = ArrayDeque()
        queue.offer(root)
        
        val result = arrayListOf<List<Int>>()
        while (queue.isNotEmpty()) {
            val levelNodes = queue.size
            val curList = arrayListOf<Int>()
            for (i in 1..levelNodes) {
                val node = queue.poll()
                curList.add(node.`val`)
                node.left?.let { queue.offer(it) }
                node.right?.let { queue.offer(it) }
            }
            result.add(curList)
        }
        
        return result
    }
}