[leetcode]99. Recover Binary Search Tree
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题目链接:https://leetcode.com/problems/recover-binary-search-tree/#/description
Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
题目有一个附加要求就是要求空间复杂度为常数空间。而算法一的空间复杂度为O(N),还不够省空间。以下的解法也是中序遍历的写法,只是非常巧妙,使用了一个prev指针。例如一颗被破坏的二叉查找树如下:
4
/ \
2 6
/ \ / \
1 5 3 7
很明显3和5颠倒了。那么在中序遍历时:当碰到第一个逆序时:为5->4,那么将n1指向5,n2指向4,注意,此时n1已经确定下来了。然后prev和root一直向后遍历,直到碰到第二个逆序时:4->3,此时将n2指向3,那么n1和n2都已经确定,只需要交换节点的值即可。prev指针用来比较中序遍历中相邻两个值的大小关系,很巧妙。
class Solution {public: void recoverTree(TreeNode* root) { help(root); swap(first->val, second->val); } void help(TreeNode* root){ if(root==NULL) return; help(root->left); if(first==NULL && prev->val >= root->val) first=prev; if(first!=NULL && prev->val >= root->val) second=root; prev=root; help(root->right); }private: TreeNode* first=NULL; TreeNode* second=NULL; TreeNode* prev = new TreeNode(INT32_MIN);};
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