99. Recover Binary Search Tree

问题描述

wo 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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解决思路

使用inorder遍历的办法，因为在二叉查询树时，inorder遍历会满足遍历到的节点是有序的，而此时为升序，所以我们只要找到两个不满足pre->val < cur->val的节点就可以了。

代码

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* first=NULL;
TreeNode* second=NULL;
TreeNode* pre = new TreeNode(INT_MIN);
void recoverTree(TreeNode* root) {
helper(root);
int tmp = first->val;
first->val = second->val;
second->val = tmp;
}
void helper(TreeNode* root) {
if (root == NULL)
return;
helper(root->left);
if (!first && pre->val >= root->val) first = pre;
if (first && pre->val >= root->val) {second = root; }
pre = root;
helper(root->right);
}
};