标签：nod lin with sibling HERE either des roo turn
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.
Example:
Input: [1,2,3,4,5]
1
/ 2 3
/ 4 5
Output: return the root of the binary tree [4,5,2,#,#,3,1]
4
/ 5 2
/ 3 1
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode upsideDownBinaryTree(TreeNode root) { if (root == null || root.left == null) { return root; } TreeNode newNode = upsideDownBinaryTree(root.left); root.left.left = root.right; root.left.right = root; root.left = null; root.right = null; return newNode; } }
[LC] 156. Binary Tree Upside Down
标签：nod lin with sibling HERE either des roo turn
原文地址：https://www.cnblogs.com/xuanlu/p/12196022.html
