Leetcode–refresh–binary Tree Upside Down

Recursive method can be Unstand easily:

1. Assume we get the sub node with this transition. So we need the current node.

2. As the symmetic, the finished sub node would replace the position of current node.

3. Old Current->left = parent->right, old current->right = parent, since we passed the these from argument. (one corner case:if parent = = NULL, there is no parent->right)

This was a button up recursive:

1 /**2 * Definition for binary tree3 * struct TreeNode {4 * int val;5 * TreeNode *left;6 * TreeNode *right;7 * TreeNode (int x): Val (x), left (null), right (null) {}8  * };9  */Ten classSolution { One  Public: ATreeNode *gettree (TreeNode *root, TreeNode *parent) { -         if(!root)returnParent//the most left of Find, but Root was NULL, so the new root is the parent. -TreeNode *current = Gettree (root->left, Root);//This is a sub node, but after conversion, it replaces the current node. theRoot->left = parent = = NULL? Null:parent->Right ; -Root->right =parent; -         returnCurrent ; -     } +TreeNode *upsidedownbinarytree (TreeNode *root) { -         returnGettree (root, NULL); +     } A};

Here's the top down iterative:

1 /**2 * Definition for binary tree3 * struct TreeNode {4 * int val;5 * TreeNode *left;6 * TreeNode *right;7 * TreeNode (int x): Val (x), left (null), right (null) {}8  * };9  */Ten classSolution { One  Public: ATreeNode *upsidedownbinarytree (TreeNode *root) { -      if(!root)returnRoot; -TreeNode *parent = NULL, *parentright =NULL; the       while(root) { -TreeNode *lnode = root->Left ; -Root->left =Parentright; -Parentright = root->Right ; +Root->right =parent; -Parent =Root; +Root =Lnode; A      } at      returnparent; -     } -};

Leetcode–refresh–binary Tree Upside Down