Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1 / 2 5 / 3 4 6The flattened tree should look like:
1 2 3 4 5 6
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); }