首页 > 代码库 > LeetCode OJ 114. Flatten Binary Tree to Linked List
LeetCode OJ 114. Flatten Binary Tree to Linked List
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.
Subscribe to see which companies asked this question
解答
先序遍历同时把节点都堆到左边,因为先序先处理左子树,所以这样操作只是更改叶节点的left而不会对遍历有影响,最后把所有left赋值给right就可以了。
/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ struct TreeNode *pre_node; void DFS(struct TreeNode *root){ if(NULL == root){ return; } if(pre_node != root){ pre_node->left = root; pre_node = root; } DFS(root->left); DFS(root->right); } void flatten(struct TreeNode* root) { pre_node = root; DFS(root); while(root != NULL){ root->right = root->left; root->left = NULL; root = root->right; } }
LeetCode OJ 114. Flatten Binary Tree to Linked List
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。