Flatten Binary Tree to Linked List (DFS)

Given a binary tree, flatten it to a linked list in-place.For example,
Given

1
/ \
2   5
/ \   \
3   4   6

The flattened tree should look like:

1
\
2
\
3
\
4
\
5
\
6

代码：

class Solution{
public:
void flatten(TreeNode *root) {
if(root==NULL) return;
TreeNode* p=root->left;
if(p==NULL){
flatten(root->right);
return;
}

while(p->right!=NULL) p=p->right;
TreeNode* temp=root->right;
root->right=root->left;
root->left=NULL;//一定不要忘记左子树要赋空
p->right=temp;

flatten(root->right);
return;

}
};

这种DFS画图最好理解了，下图是我的解题过程：