leetcode--Recover Binary Search Tree*

1.题目描述

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?

2.解法分析

对有异常的BST进行pre-order的遍历，遍历过程中第一次发现相邻元素的逆序，即前一个元素值比后一个节点的元素值大，那么前一个节点是被交换的节点中的其中一个，记为sw1，继续遍历，找到第一个比sw1中元素值大的节点sw2,如果找到，sw1和sw2，交换它们的元素值即可。细节参考代码。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<TreeNode *> v;
        TreeNode * curTN=root;
        TreeNode *prevTN=NULL;
        TreeNode *sw1=NULL;
        TreeNode *sw2=NULL;

        while(!v.empty()||curTN)
        {
            while(curTN)
            {
                v.push_back(curTN);
                curTN=curTN->left;
            }
            if(!v.empty())
            {
                curTN =v.back();v.pop_back();
                if(!sw1){
                    if(!prevTN){
                        prevTN=curTN;
                    }
                    else
                    {
                        if(curTN->val<prevTN->val){
                            sw1=prevTN;
                        }
                        prevTN =curTN;
                    }
                }
                else
                {
                    if(curTN->val>sw1->val){sw2=prevTN;break;}
                    else prevTN=curTN;
                }

                curTN = curTN->right;
            }
        }

        if(!sw1)return; //未找到异常交换
        if(!sw2)sw2=prevTN;//最大的一个节点被交换
        int temp=sw1->val;
        sw1->val=sw2->val;
        sw2->val=temp;

    }
};

另外题目说不用O(N)的空间复杂度会有加分，在discuss里面找了一个，没仔细看，思路大概跟我的一样，但是就是记录路径可能用了不同的办法，貌似叫做morris traversal，代码如下，做个记录，之后再分析（今天写了一篇文章，详细介绍了Morris Traveral)：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode *h) {
    TreeNode *f1, *f2;
    bool found = false;
    TreeNode *pre, *par = 0; // previous AND parent
    while(h) { // Morris Traversal
        if(h->left == 0) {
            if(par && par->val > h->val) {  // inorder previous is: par
                if(!found) {
                    f1 = par;
                    found = true;
                }
                f2 = h;
            }
            par = h;
            h = h->right;
            continue;
        }
        pre = h->left;
        while(pre->right != 0 && pre->right != h)
            pre = pre->right;
        if(pre->right == 0) {
            pre->right = h;
            h = h->left;
        } else {
            pre->right = 0;
            if(pre->val > h->val) { // inorder previous is: pre
                if(!found) {
                    f1 = pre;
                    found =true;
                }
                f2 = h;
            }
            par = h;
            h = h->right;
        }
    }
    if(found)
        swap(f1->val, f2->val);
}
};