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[Swift]LeetCode156.二叉树的上下颠倒 $ Binary Tree Upside Down

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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 }

 

posted @ 2018-12-03 19:32  为敢技术  阅读(251)  评论(0编辑  收藏  举报