【Leetcode】99. Recover Binary Search Tree（二叉树两个错误位置交换）

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

Recover the tree without changing its structure.

Example 1:

Input: [1,3,null,null,2]

   1
  /
 3
  \
   2

Output: [3,1,null,null,2]

   3
  /
 1
  \
   2

Example 2:

Input: [3,1,4,null,null,2]

  3
 / \
1   4
   /
  2

Output: [2,1,4,null,null,3]

  2
 / \
1   4
   /
  3

Follow up:

• A solution using O(n) space is pretty straight forward.
• Could you devise a constant space solution?

题目大意：

二叉树中有两个错误位置，交换这两个位置。

解题思路：

中序遍历之后，sort排序可以得到正确的二叉树。此时再次中序遍历，找到位置不对应的两个位置返回，交换中间值。

/**
 * 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 {
private:
    vector<int> ans;
    int idx ;
    void search(TreeNode* tmp){
        if(tmp->left==NULL&&tmp->right==NULL){
            ans.push_back(tmp->val);
            return;
        }
        if(tmp->left!=NULL)
            search(tmp->left);
        ans.push_back(tmp->val);
        if(tmp->right!=NULL)
            search(tmp->right);
    }
    void search_to(TreeNode *tmp, TreeNode *&node1, TreeNode *&node2){
        if(tmp->left==NULL&&tmp->right==NULL){
            if(tmp->val != ans[idx]){
                if(node1 == NULL){
                    node1 = tmp;
                }else{
                    node2 = tmp;
                }
            }
            idx++;
            return;
        }
        if(tmp->left!=NULL)
            search_to(tmp->left, node1, node2);
        if(tmp->val != ans[idx]){
            if(node1 == NULL){
                node1 = tmp;
            }else{
                node2 = tmp;
            }
        }
        idx++;
        if(tmp->right!=NULL)
            search_to(tmp->right, node1, node2);
    }
public:
    void recoverTree(TreeNode* root) {
        // 中序遍历->vector
        // 排序
        // vector->中序替换
        search(root);
        int len = ans.size();
        sort(ans.begin(), ans.end());
        idx = 0;
        TreeNode *node1=NULL, *node2=NULL;
        search_to(root, node1, node2);
        int val = node1->val;
        node1->val = node2->val;
        node2->val = val;
    }
};