Leetcode: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
Hints:
If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.
分析:这道题是pre-order traversal的变种,我们可以很容易的写出recursion的解法,先flatten左子树再flatten右子树,然后将左子树最右节点的right child设为右子树的根。但空间复杂度为O(logn)。代码如下:
class Solution { public: void flatten(TreeNode *root) { if(root == NULL) return; flatten(root->left); flatten(root->right); if(root->left == NULL) return; TreeNode *p = root->left; for(; p->right; p = p->right);//find right most node in left child tree p->right = root->right; root->right = root->left; root->left = NULL; } };
如果要用in place的方法,我们可以联想到Morris遍历。代码如下:
class Solution { public: void flatten(TreeNode *root) { TreeNode *cur = root; while(cur){ if(cur->left != NULL){ TreeNode *prev = cur->left; for(; prev->right && prev->right != cur; prev = prev->right); if(prev->right == NULL){ prev->right = cur; cur = cur->left; }else{ prev->right = cur->right; cur->right = cur->left; cur->left = NULL; cur = prev->right; } }else{ cur = cur->right; } } } };