@@ -646,5 +646,68 @@ func countNodes(_ root: TreeNode?) -> Int {
646646}
647647```
648648
649+ ## Scala
650+ 651+ 递归:
652+ ``` scala
653+ object Solution {
654+ def countNodes (root : TreeNode ): Int = {
655+ if (root == null ) return 0
656+ 1 + countNodes(root.left) + countNodes(root.right)
657+ }
658+ }
659+ ```
660+ 661+ 层序遍历:
662+ ``` scala
663+ object Solution {
664+ import scala .collection .mutable
665+ def countNodes (root : TreeNode ): Int = {
666+ if (root == null ) return 0
667+ val queue = mutable.Queue [TreeNode ]()
668+ var node = 0
669+ queue.enqueue(root)
670+ while (! queue.isEmpty) {
671+ val len = queue.size
672+ for (i <- 0 until len) {
673+ node += 1
674+ val curNode = queue.dequeue()
675+ if (curNode.left != null ) queue.enqueue(curNode.left)
676+ if (curNode.right != null ) queue.enqueue(curNode.right)
677+ }
678+ }
679+ node
680+ }
681+ }
682+ ```
683+ 684+ 利用完全二叉树性质:
685+ ``` scala
686+ object Solution {
687+ def countNodes (root : TreeNode ): Int = {
688+ if (root == null ) return 0
689+ var leftNode = root.left
690+ var rightNode = root.right
691+ // 向左向右往下探
692+ var leftDepth = 0
693+ while (leftNode != null ) {
694+ leftDepth += 1
695+ leftNode = leftNode.left
696+ }
697+ var rightDepth = 0
698+ while (rightNode != null ) {
699+ rightDepth += 1
700+ rightNode = rightNode.right
701+ }
702+ // 如果相等就是一个满二叉树
703+ if (leftDepth == rightDepth) {
704+ return (2 << leftDepth) - 1
705+ }
706+ // 如果不相等就不是一个完全二叉树,继续向下递归
707+ countNodes(root.left) + countNodes(root.right) + 1
708+ }
709+ }
710+ ```
711+ 649712-----------------------
650713<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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