[Locked] Binary Tree Upside Down

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.

For 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  

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

分析：

　　自底向上的右旋，用DFS搜到最左下角的节点，然后依次进行处理

代码：

class Solution {
private:
    TreeNode *newRoot;
public:
    void dfs(TreeNode* node) {
        if(!node->left) {
            newRoot = node;
            return;
        }
        dfs(node->left);
        node->left->left = node->right;
        node->left->right = node;
        return;
    }
    TreeNode* upsideDown(TreeNode* root) {
        if(root)
            dfs(root);
        return newRoot;
    }
};