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

我就是我 2022-04-24 03:30 130阅读 0赞

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;
        }
    };