Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?
思路：
如果不限定空间的话，直接的办法是中序遍历的值放入一个数组，然后遍历这个数组，找到第一个前面大于后面的数的较大的数，再找最后一个前面大于后面的数的较小的数。因为要求是固定空间，所以用深搜实现，用pre和cur分别记录前一个和当前一个节点，如果pre>cur的话则记录错误节点，最后把错误节点的数值互换。
代码：
1 TreeNode *pre, *cur, *n1, *n2; 2 void search(TreeNode *root){ 3 if(!root) 4 return; 5 search(root->left); 6 cur = root; 7 if(pre && pre->val > cur->val){ 8 if(!n1){ 9 n1 = pre; 10 n2 = cur; 11 } 12 else{ 13 n2 = cur; 14 } 15 } 16 pre = root; 17 search(root->right); 18 } 19 void recoverTree(TreeNode *root) { 20 // IMPORTANT: Please reset any member data you declared, as 21 // the same Solution instance will be reused for each test case. 22 if(!root) 23 return; 24 pre = NULL; 25 cur = NULL; 26 n1 = NULL; 27 n2 = NULL; 28 search(root); 29 int tmp = n1->val; 30 n1->val = n2->val; 31 n2->val = tmp; 32 }
第二遍传指针参数进去结果出了些问题，后来改成**好了。传参的概念有时候还是不太清晰。
1 void search(TreeNode *root, TreeNode **pre, TreeNode **cur, TreeNode **n1, TreeNode **n2){ 2 if(root){ 3 search(root->left, pre, cur, n1, n2); 4 *cur = root; 5 if(*pre && (*pre)->val > (*cur)->val){ 6 if(!*n1){ 7 *n1 = *pre; 8 } 9 *n2 = *cur; 10 } 11 *pre = root; 12 search(root->right, pre, cur, n1, n2); 13 } 14 } 15 void recoverTree(TreeNode *root) { 16 // IMPORTANT: Please reset any member data you declared, as 17 // the same Solution instance will be reused for each test case. 18 TreeNode *pre = NULL, *cur = NULL, *n1 = NULL, *n2 = NULL; 19 search(root, &pre, &cur, &n1, &n2); 20 if(n1 && n2){ 21 int tmp = n1->val; 22 n1->val = n2->val; 23 n2->val = tmp; 24 } 25 }