Flatten Binary Tree to Linked List
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题目:
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
思想:
有上述的描述可知,转换后的链表的连接顺序为二叉树先序便利的顺序,为此,我们可以采用先序遍历的思想,先递归的处理左子树将其转换为一个链表,然后再处理右子树将其转换为一个链表,但是注意先连接左子树的结果再连接右子树的结果。
代码:
class Solution {public:void flatten(TreeNode* root) {if (root == NULL){return ;}flatten(root->left);flatten(root->right);TreeNode *tempR = root->right;root->right = root->left;root->left = nullptr;while (root->right){root = root->right;}root->right= tempR;}};
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