@@ -546,7 +546,50 @@ func findBottomLeftValue(_ root: TreeNode?) -> Int {
546546}
547547```
548548
549+ ## Scala
550+ 551+ 递归版本:
552+ ``` scala
553+ object Solution {
554+ def findBottomLeftValue (root : TreeNode ): Int = {
555+ var maxLeftValue = 0
556+ var maxLen = Int .MinValue
557+ // 递归方法
558+ def traversal (node : TreeNode , leftLen : Int ): Unit = {
559+ // 如果左右都为空并且当前深度大于最大深度,记录最左节点的值
560+ if (node.left == null && node.right == null && leftLen > maxLen) {
561+ maxLen = leftLen
562+ maxLeftValue = node.value
563+ }
564+ if (node.left != null ) traversal(node.left, leftLen + 1 )
565+ if (node.right != null ) traversal(node.right, leftLen + 1 )
566+ }
567+ traversal(root, 0 ) // 调用方法
568+ maxLeftValue // return关键字可以省略
569+ }
570+ }
571+ ```
549572
573+ 层序遍历:
574+ ``` scala
575+ import scala .collection .mutable
576+ 577+ def findBottomLeftValue (root : TreeNode ): Int = {
578+ val queue = mutable.Queue [TreeNode ]()
579+ queue.enqueue(root)
580+ var res = 0 // 记录每层最左侧结果
581+ while (! queue.isEmpty) {
582+ val len = queue.size
583+ for (i <- 0 until len) {
584+ val curNode = queue.dequeue()
585+ if (i == 0 ) res = curNode.value // 记录最最左侧的节点
586+ if (curNode.left != null ) queue.enqueue(curNode.left)
587+ if (curNode.right != null ) queue.enqueue(curNode.right)
588+ }
589+ }
590+ res // 最终返回结果,return关键字可以省略
591+ }
592+ ```
550593
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552595<div align =" center " ><img src =https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width =500 > </img ></div >
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