99. Recover Binary Search Tre

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?

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解法:来自leetcode 150题集

O(n)的解法,开一个指针数组,中序遍历,将节点指针一次存放在数组中,

然后寻找两处逆向的位置,先从前往后找第一个逆序的位置,然后从后往前寻找第二个逆序的位置,交换两个指针的值,

递归/非递归中序遍历一般需要用到栈,空间也是O(n)的,可以利用morris中序遍历的方式

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code:

/**
 * 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){
        pair<TreeNode *,TreeNode *> broken;
        TreeNode *curr = root;
        TreeNode *prev = nullptr;
        broken.first = broken.second = nullptr;
 
        while(curr!=nullptr){
            if(curr->left==nullptr){
                detect(broken,prev,curr);
                prev = curr;
                curr = curr->right;
            }else{
                auto node = curr->left;
                ///prev = curr->left;
                while(node->right != nullptr && node->right!=curr){
                    node = node->right;
                }
 
                ///find predecessor
                if(node->right==nullptr){
                    node->right = curr;
                    curr = curr->left;
                }else{
                    node->right = nullptr;
                    detect(broken,prev,curr);
                    prev = curr;
                    curr = curr->right;
                }
            }///if-else
        }///while
 
        swap(broken.first->val,broken.second->val);
    }
    void detect(pair<TreeNode *,TreeNode *> &broken,TreeNode *prev,
        TreeNode *curr){
        if(prev!=nullptr && prev->val > curr->val){
            if(broken.first == nullptr) broken.first = prev;
            broken.second = curr;
        }
    }
};