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：