题目描述

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.

样例

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  

算法1

(枚举) $O(n)$

题意是当前node与左节点的父子关系颠倒，node的右节点变为node左节点的左节点。所以需要四个指针：cur指向当前节点，pre指向父节点，tmp指向右节点，next指向左节点。从next指针开始迭代。

时间复杂度

遍历一次O(n)
空间O（1）

参考文献

C++ 代码

class Solution {
public:
    TreeNode* upsideDownBinaryTree(TreeNode* root) {
        if(!root) return NULL;
        TreeNode *cur=root, *pre=NULL, *next=NULL, *tmp=NULL;
        while(cur)
        {
            next=cur->left;
            cur->left=tmp;
            tmp=cur->right;
            cur->right=pre;
            pre=cur;
            cur=next;
        }
        return pre;
    }
};