leetcode_c++:树: Recover Binary Search Tree(099)

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?

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//有一个BST有两个节点不小心互换了，恢复这颗树

算法

中序遍历，找出节点，互换

时间：O(n)

空间：o(n)

class Solution {
public:
TreeNode *p,*q;
TreeNode *prev;
void recoverTree(TreeNode *root)
{
p=q=prev=NULL;
inorder(root);
swap(p->val,q->val);
}
void inorder(TreeNode *root)
{
if(root->left)inorder(root->left);
if(prev!=NULL&&(prev->val>root->val))
{
if(p==NULL)p=prev;
q=root;
}
prev=root;
if(root->right)inorder(root->right);
}
};

算法

O(1)的解法

http://fisherlei.blogspot.kr/2012/12/leetcode-recover-binary-search-tree.html

这里写代码片