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99. Recover Binary Search Tree

2024-09-05 07:21:42   216人阅读

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?

解题思路：本题主要是考查了Morris Traversal，一种根据线索二叉树思想的时间复杂度为O(n),空间复杂度为O(1)的中序遍历方式。

在博客http://www.cnblogs.com/AnnieKim/archive/2013/06/15/MorrisTraversal.html里，原作者对Morris Traversal说的很明白。此题在此基础上只要比较相邻两个数的大小，就可以确定哪两个是打乱序的了。而Morris Traversal的中序输出就是原数组的从小到大排序，因而一切就变得简单了

/**
 * 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) {
        TreeNode *pre=NULL,*cur=root;
        TreeNode *first=NULL,*second=NULL,*temp=NULL;
        while(cur!=NULL){
            if(cur->left==NULL){
                //cur->val;
                if(temp!=NULL&&temp->val>cur->val){
                    if(!first)first=temp;
                    second=cur;
                }
                temp=cur;
                cur=cur->right;
            }
            else {
                pre=cur->left;
                while(pre->right!=NULL&&pre->right!=cur)
                pre=pre->right;
                //exist threaded tree
                if(pre->right==cur){
                    //print cur->val;
                    if(temp!=NULL&&temp->val>cur->val){
                        if(!first)first=temp;
                        second=cur;
                    }
                    temp=cur;
                    pre->right=NULL;
                    cur=cur->right;
                }
                //creating backend poinner to the current point
                else{
                    pre->right=cur;
                    cur=cur->left;
                }
            }
        }
        swap(first->val,second->val);
    }
};

 

99. Recover Binary Search Tree


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