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
分析：
对于树，递归总是是很好的实现方法，所以这题应该
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public TreeNode upsideDownBinaryTree(TreeNode root) { 12 if (root == null || root.left == null) { 13 return root; 14 } 15 TreeNode newRoot = upsideDownBinaryTree(root.left); 16 17 root.left.right = root; 18 root.left.left = root.right; 19 20 root.left = null; 21 root.right = null; 22 23 return newRoot; 24 } 25 }
原文：http://www.cnblogs.com/beiyeqingteng/p/6064106.html