156. Binary Tree Upside Down

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  

Clarification:

Confused what [4,5,2,#,#,3,1] means? Read more below on how binary tree is serialized on OJ.

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as [1,2,3,#,#,4,#,#,5].

這題有點像是每個root, left child, right child順時針旋轉。

// Recursion Method
TreeNode* upsideDownBinaryTree(TreeNode* root) { // time: O(n); space: O(n)
    if (!root || !root->left) return root;
    TreeNode *l = root->left, *r = root->right;
    TreeNode *newRoot = upsideDownBinaryTree(l);
    l->left = r; // original right node becomes left node
    l->right = root; // original root becomes right node
    root->left = nullptr;
    root->right = nullptr;
    return newRoot;
}

next: 當前這輪的left node是下一輪的current node pre: 當前這輪的current node是下一輪的right node tmp: 當前這輪的right node是下一輪的left node

// Iteration Method
TreeNode* upsideDownBinaryTree(TreeNode* root) { // time: O(n); space: O(1)
    TreeNode *cur = root, *next = nullptr, *pre = nullptr, *tmp = nullptr;
    while (cur) {
        next = cur->left; // the next iteration cur node
        cur->left = tmp;
        tmp = cur->right; // current right node is left node in the next iteration
        cur->right = pre;
        pre = cur; // cur root node is the right node in the next iteration
        cur = next;
    }
    return pre;
}