156.Binary Tree Upside Down

Tags: Tree Medium

Links: https://leetcode-cn.com/problems/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].

---

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* upsideDownBinaryTree(TreeNode* root) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        if (!root) return NULL;
        if (!root -> left && !root -> right) return root;

        TreeNode *newRoot = upsideDownBinaryTree(root -> left);
        root -> left -> left = root -> right;
        root -> left -> right = root;
        root -> left = root -> right = NULL;

        return newRoot;
    }
};

给定一棵二叉树，所有右节点要么是空，要么是叶节点加左兄弟节点。 通俗来讲，这棵二叉树是一条左链，上面挂着若干右儿子，如下所示：

现在请将左链自下到上翻转成右链，并将所有右儿子换到左侧。