题目如下:
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.
分析如下:
我非常喜欢这道题目的转换的思路。
第一个办法是改造in order遍历,这个和 Convert Sorted List to Binary Search Tree 表面上完全不同,实际上核心一致,都是根据需要对树的中序遍历进行修改。对普通的中序遍历递归版进行更改,就可以解决这道题目。这个变换可以无穷多。
第二个办法是 Morris遍历,其实是对树的线索化。让叶节点的右孩子指向它的中序遍历的后继,结束遍历之后恢复叶节点的右孩子为NULL。 第二个办法的实现留在以后来完成。
我的代码:
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/*
* 解法1 改造in order遍历 http://yucoding.blogspot.com/2013/03/leetcode-question-75-recover-binary.html
* 解法2 使用morris遍历 1.http://chuansongme.com/n/100461 2.http://www.cnblogs.com/TenosDoIt/p/3445682.html
*/
class Solution { //解法1
public:
void recoverTree(TreeNode *root, TreeNode* & pre, TreeNode* & first, TreeNode* & second) {
if (root == NULL) return;
recoverTree(root->left, pre, first, second);
if (pre == NULL) {
pre = root;
}else {
if (pre->val > root->val) {
if (first == NULL)
first = pre;
second = root;
}
pre = root;
}
recoverTree(root->right, pre, first, second);
}
void recoverTree(TreeNode *root) {
if (root == NULL) return;
TreeNode* first = NULL;
TreeNode* second = NULL;
TreeNode* pre = NULL;
recoverTree(root, pre, first, second);
int tmp = first->val;
first->val = second->val;
second->val = tmp;
return;
}
};
参考资料: