Jun 13, 2022 · May 19, 2022 · May 19, 2022 · May 19, 2022 · May 19, 2022 · May 19, 2022
diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md
diff --git a/problems/0226.翻转二叉树.md b/problems/0226.翻转二叉树.md
Original file line number	Diff line number	Diff line change
Expand Up		@@ -725,5 +725,25 @@ func isSymmetric3(_ root: TreeNode?) -> Bool {
		}
		```

	## Scala

	递归:
	```scala
	object Solution {
	def isSymmetric(root: TreeNode): Boolean = {
	if (root == null) return true // 如果等于空直接返回true
	def compare(left: TreeNode, right: TreeNode): Boolean = {
	if (left == null && right == null) return true // 如果左右都为空,则为true
	if (left == null && right != null) return false // 如果左空右不空,不对称,返回false
	if (left != null && right == null) return false // 如果左不空右空,不对称,返回false
	// 如果左右的值相等,并且往下递归
	left.value == right.value && compare(left.left, right.right) && compare(left.right, right.left)
	}
	// 分别比较左子树和右子树
	compare(root.left, root.right)
	}
	}
	```

		-----------------------
		<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
Original file line number	Diff line number	Diff line change
Expand Up		@@ -818,5 +818,53 @@ func invertTree(_ root: TreeNode?) -> TreeNode? {
		}
		```

	### Scala

	深度优先遍历(前序遍历):
	```scala
	object Solution {
	def invertTree(root: TreeNode): TreeNode = {
	if (root == null) return root
	// 递归
	def process(node: TreeNode): Unit = {
	if (node == null) return
	// 翻转节点
	val curNode = node.left
	node.left = node.right
	node.right = curNode
	process(node.left)
	process(node.right)
	}
	process(root)
	root
	}
	}
	```

	广度优先遍历(层序遍历):
	```scala
	object Solution {
	import scala.collection.mutable
	def invertTree(root: TreeNode): TreeNode = {
	if (root == null) return root
	val queue = mutable.Queue[TreeNode]()
	queue.enqueue(root)
	while (!queue.isEmpty) {
	val len = queue.size
	for (i <- 0 until len) {
	var curNode = queue.dequeue()
	if (curNode.left != null) queue.enqueue(curNode.left)
	if (curNode.right != null) queue.enqueue(curNode.right)
	// 翻转
	var tmpNode = curNode.left
	curNode.left = curNode.right
	curNode.right = tmpNode
	}
	}
	root
	}
	}
	```

		-----------------------
		<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>