leetcode dfs Flatten Binary Tree to Linked List

Flatten Binary Tree to Linked List

 Total Accepted: 25034 Total Submissions: 88947My Submissions

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

click to show hints.

Hints:

If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

题意：把一棵二叉树变为链表，要求原地进行转变。
思路：dfs
能够注意到。在链表表示中，每一个节点的下一个节点是二叉树的先序遍历中的下一个节点。所以问题就转换为先序遍历了。
复杂度：时间O(n)，空间O(log n)

TreeNode *cur;
void flatten(TreeNode *root) {
	if(!root) return;
	TreeNode *left = root->left;
	TreeNode *right = root->right;
	if(cur){
		cur->left = NULL;
		cur->right = root;
	}
	cur = root;
	flatten(left);
	flatten(right);
}