Binary Tree Upside Down解题报告

Description:

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:

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  

Link:

https://leetcode.com/problems/binary-tree-upside-down/description/

解题方法：

很恶心的一道题，之前花了大量时间在序列化和反序列化上来解这道题，后来发现其实不需要这样做。
这颗树是一颗左偏的树，而题目要求将这棵树上下颠倒并且变成右偏，那么在简单的数结构：

   1
  / \
 2   3
或者是
   1
  / \
 2   null

之中，上下颠倒代表着子节点之中一个要成为父节点。
因为结果是一颗右偏的数，所以原右孩子会成为之后的左孩子（因为原右孩子和变形之后的左孩子一样，不是叶子就是NULL）；而原左孩子则成为父节点；而原父节点毫无疑问只能成为右子树：

   2
  / \
 3   1
或者是
   2
  /   \
 null  1

所以我们可以用迭代来完成这样的变形，只要记录下每层变形之后的父节点即可。

Time Complexity：

O(N)

完整代码：

TreeNode* upsideDownBinaryTree(TreeNode* root) 
    {
        TreeNode* parent = root;
        if(!parent)
            return root;
        TreeNode* left = parent->left;
        TreeNode* right = parent->right;
        parent->left = NULL;
        parent->right = NULL;
        //when left == NULL which means we touch the most left leaf
        while(left)  
        {
            TreeNode* nextLeft = left->left;
            TreeNode* nextRight = left->right;
            left->left = right;
            left->right = parent;
            //restore the parent node after one switch
            parent = left; 
            left = nextLeft;
            right = nextRight;
        }
        return parent;
    }