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Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.
For example: Given a binary tree {1,2,3,4,5},
1
/
2 3
/
4 5
return the root of the binary tree [4,5,2,#,#,3,1].
4
/
5 2
/
3 1
给定一个二叉树，其中所有右节点要么是具有兄弟节点的叶节点（共享同一父节点的左节点），要么是空的，将其颠倒并将其转换为树，其中原始右节点转换为左叶节点。返回新根。
例如：给定二叉树{1，2，3，4，5 }，
1
/
2 3
/
4 5
返回二叉树的根 [4,5,2,#,#,3,1].
4
/
5 2
/
3 1
Solution：
1 public class TreeNode { 2 public var val: Int 3 public var left: TreeNode? 4 public var right: TreeNode? 5 public init(_ val: Int) { 6 self.val = val 7 self.left = nil 8 self.right = nil 9 } 10 } 11 12 class Solution { 13 func upsideDownBinaryTree(_ root: TreeNode?) -> TreeNode? { 14 if root == nil || root!.left == nil {return root} 15 var l: TreeNode? = root!.left 16 var r: TreeNode? = root!.right 17 var res: TreeNode? = upsideDownBinaryTree(l) 18 l!.left = r 19 l!.right = root 20 root!.left = nil 21 root!.right = nil 22 return res 23 } 24 }