[leetcode]Recover Binary Search Tree

Recover Binary Search Tree

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

题意:交换了二叉排序树中的两个节点，要求将其还原，不改变树的结构。

思路:中序遍历二叉树,如果pre->val > cur->val，那么将这两个节点存在一个容器里，中序遍历之后，如果容器中有两个元素，说明那两个节点相邻，交换这两个节点；如果容器中有四个元素，则将第一个和第四个交换即可。

/**
 * 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 swap(int &a,int &b)
    {
        int t = a;
        a = b;
        b = t;
    }
    TreeNode *pre;
    void inorder(TreeNode *cur,vector<TreeNode*> &q)
    {
        if(!cur)    return;
        inorder(cur->left,q);
        if(pre && pre->val > cur->val){
            q.push_back(pre);
            q.push_back(cur);
            
        }
        pre = cur;
        inorder(cur->right,q); 
    }
    void recoverTree(TreeNode* root) {
        vector<TreeNode*> q;
        pre = NULL;
        inorder(root,q);
       
        if(q.size() == 2)
            swap(q[0]->val,q[1]->val);
        else if(q.size() == 4)
        {
            swap(q[0]->val,q[3]->val);
        }
        
    }
};