Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Example 1:
Input: [1,3,null,null,2]
1
/
3
\
2
Output: [3,1,null,null,2]
3
/
1
\
2
Example 2:
Input: [3,1,4,null,null,2]
3
/ \
1 4
/
2
Output: [2,1,4,null,null,3]
2
/ \
1 4
/
3
Follow up:
- A solution using O(n) space is pretty straight forward.
- Could you devise a constant space solution?
解法
本来想了一些稍复杂的算法,后来发现完全不需要,利用二叉搜索树的排序性质,中序遍历即可得到排序的序列。先一个遍历获取到序列,然后对序列排序,最后再来一个中序遍历重构二叉树(只需要该节点值即可)
/**
* 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 {
int i;
void InOrderVis(TreeNode* root, vector<int>& node) {
if(root) {
InOrderVis(root->left, node);
node.push_back(root->val);
InOrderVis(root->right, node);
}
}
void InOrderVisBuild(TreeNode* root, vector<int>& node) {
if(root) {
InOrderVisBuild(root->left, node);
root->val = node[i++];
InOrderVisBuild(root->right, node);
}
}
public:
void recoverTree(TreeNode* root) {
vector<int> node;
InOrderVis(root, node);
sort(node.begin(), node.end());
i=0;
InOrderVisBuild(root, node);
}
};