添加 0222.完全二叉树的节点个数.md Scala版本

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ZongqinWang
2022-05-23 19:14:45 +08:00
parent b78e750f8f
commit f83d5edb6e

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@ -646,5 +646,68 @@ func countNodes(_ root: TreeNode?) -> Int {
}
```
## Scala
递归:
```scala
object Solution {
def countNodes(root: TreeNode): Int = {
if(root == null) return 0
1 + countNodes(root.left) + countNodes(root.right)
}
}
```
层序遍历:
```scala
object Solution {
import scala.collection.mutable
def countNodes(root: TreeNode): Int = {
if (root == null) return 0
val queue = mutable.Queue[TreeNode]()
var node = 0
queue.enqueue(root)
while (!queue.isEmpty) {
val len = queue.size
for (i <- 0 until len) {
node += 1
val curNode = queue.dequeue()
if (curNode.left != null) queue.enqueue(curNode.left)
if (curNode.right != null) queue.enqueue(curNode.right)
}
}
node
}
}
```
利用完全二叉树性质:
```scala
object Solution {
def countNodes(root: TreeNode): Int = {
if (root == null) return 0
var leftNode = root.left
var rightNode = root.right
// 向左向右往下探
var leftDepth = 0
while (leftNode != null) {
leftDepth += 1
leftNode = leftNode.left
}
var rightDepth = 0
while (rightNode != null) {
rightDepth += 1
rightNode = rightNode.right
}
// 如果相等就是一个满二叉树
if (leftDepth == rightDepth) {
return (2 << leftDepth) - 1
}
// 如果不相等就不是一个完全二叉树,继续向下递归
countNodes(root.left) + countNodes(root.right) + 1
}
}
```
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>