leetcode:Recover Binary Search Tree
2014-10-29 15:39
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又有好几天没有做题,这道题看了题目很久了,一直在思考,因为要求常数空间而实际上递归解法需要的空间并不是常数级别的。看到网上的解答,有人说可以用Morris方法遍历二叉树,这样可以在o(1)空间复杂度和o(n)时间复杂度完成二叉树的遍历。但是好几天没有刷题了,想要快点做出这道题,所以还是用了递归的解法。Morris方法下次再做吧~
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
read more on how binary tree is serialized on OJ.
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题目地址:https://oj.leetcode.com/problems/recover-binary-search-tree/
解题思路:思路是先得到二叉树的中序遍历,找出顺序不对的数,如果只有一个地方顺序不对交换两个数;如果出现两个顺序不对的地方,把第一个地方的较大的数和第二个地方较小的数交换。用change这个vector来存储顺序不对的节点。不管最后change中是只有两个节点还是有4个节点,都是交换第一个节点和最后一个节点的值。
代码:
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.
Show Tags
Have you met this question in a real interview?
Yes
No
题目地址:https://oj.leetcode.com/problems/recover-binary-search-tree/
解题思路:思路是先得到二叉树的中序遍历,找出顺序不对的数,如果只有一个地方顺序不对交换两个数;如果出现两个顺序不对的地方,把第一个地方的较大的数和第二个地方较小的数交换。用change这个vector来存储顺序不对的节点。不管最后change中是只有两个节点还是有4个节点,都是交换第一个节点和最后一个节点的值。
代码:
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode *pre; vector<TreeNode *> change; void recoverTree(TreeNode *root) { //思路是先得到二叉树的中序遍历,找出顺序不对的数,如果只有一个地方顺序不对交换两个数;如果出现两个顺序不对的地方,把第一个地方的较大的数和第二个地方较小的数交换 if(root==NULL) return; if(root->left==NULL&&root->right==NULL) return; pre=NULL; change.clear(); inorder(root); //交换change第一个节点和最后一个节点的值 if(change.size()>0){ int tmp=change[0]->val; change[0]->val=change[change.size()-1]->val; change[change.size()-1]->val=tmp; } } void inorder(TreeNode *root){ if(root == NULL) return; inorder(root->left); if(pre == NULL){ pre = root; }else{ if(pre->val>root->val){ change.push_back(pre); change.push_back(root); } pre = root; } inorder(root->right); } };
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