思路:按照中序遍历将root遍历一遍,并将每个位置的地址和该节点的值分别保存到数组中。然后将保存节点值的数组从小到大排序,按地址的顺序赋给每个节点即可。
/**
* 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 {
public:
void recoverTree(TreeNode* root) {
vector<TreeNode*> list;
vector<int> value;
inorder(root,list,value);
sort(value.begin(),value.end());
for(int i=0;i<value.size();++i)
{
list[i]->val=value[i];
}
}
void inorder(TreeNode* root,vector<TreeNode*> &list,vector<int> &value)
{
if(!root) return ;
inorder(root->left,list,value);
list.push_back(root);
value.push_back(root->val);
inorder(root->right,list,value);
}
};