2015-10-18 10:42:52 +08:00

91 lines
2.3 KiB
C++

// Source : https://oj.leetcode.com/problems/recover-binary-search-tree/
// Author : Hao Chen
// Date : 2014-10-11
/**********************************************************************************
*
* 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?
*
* confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
*
* OJ's Binary Tree Serialization:
*
* The serialization of a binary tree follows a level order traversal, where '#' signifies
* a path terminator where no node exists below.
*
* Here's an example:
*
* 1
* / \
* 2 3
* /
* 4
* \
* 5
*
* The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
*
*
**********************************************************************************/
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
//
// We can convert the BST to a sorted array, then we can find the two nodes which missed the order.
//
// To cover the BST to sorted array, we needn't use an extra array, we just traverse the tree in order.
//
// 8
// _______/ \_______
// / \
// 4 12
// __/ \__ __/ \__
// / \ / \
// 2 6 10 14
// / \ / \ / \ / \
// 1 3 5 7 9 11 13 15
//
//
class Solution {
public:
void recoverTreeHelper(TreeNode *root) {
if (root == NULL) return;
recoverTreeHelper(root->left);
if (prev) {
if (prev->val > root->val){
if (n1==NULL) {
n1 = prev;
}
n2 = root;
}
}
prev = root;
recoverTreeHelper(root->right);
}
void recoverTree(TreeNode *root) {
n1 = n2 = prev = NULL;
recoverTreeHelper(root);
if (n1 && n2) {
swap(n1->val, n2->val);
}
}
private:
TreeNode *n1, *n2, *prev;
};