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
思路：不是特别理解怎么翻转，用recursive做的。完全按照题目给的例子来。翻转左子树，然后原来的root.left重新分配指向。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public TreeNode upsideDownBinaryTree(TreeNode root) { if(root==null||root.left==null&&root.right==null) { return root; } TreeNode check=upsideDownBinaryTree(root.left); root.left.left=root.right; root.left.right=root; root.left=null; root.right=null; return check; } }
原文：http://www.cnblogs.com/Machelsky/p/5894786.html