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：