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99. Recover Binary Search Tree

时间:2020-12-09 12:24:32      阅读:5      评论:0      收藏:0      [点我收藏+]

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问题:

给定一颗BST(二叉搜索树),其中存在两个节点顺序颠倒,请找出他们,并恢复正确次序。

Example 1:
Input: root = [1,3,null,null,2]
Output: [3,1,null,null,2]
Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid.

Example 2:
Input: root = [3,1,4,null,null,2]
Output: [2,1,4,null,null,3]
Explanation: 2 cannot be in the right subtree of 3 because 2 < 3. Swapping 2 and 3 makes the BST valid.
 

Constraints:
The number of nodes in the tree is in the range [2, 1000].
-231 <= Node.val <= 231 - 1

技术图片        技术图片

 

 

解法:Binary Search Tree(二叉搜索树)inorder traverse(中序遍历)

由于是BST,

可知,中序遍历+排序(从小到大)-> BST

因此,我们同样通过中序遍历,来寻找有问题的两个节点。

 

步骤:

1. 递归中序遍历二叉树

2. 找到问题节点first和second

  1. 对于每个当前节点node,我们需要判定:如果 node->val < pre->val ,那么当前节点和pre节点,一定存在一个节点不正确。

                     (pre节点为遍历的前一个节点,每次处理完毕更新pre=当前node)

  2. 而我们要找的first和second,具有如下特点:

    1. first:最左边不正确的节点。

        而对于中序遍历过程中,第一次保存的有问题的pre,则为first。

    2. second:最右边不正确的节点。
        而对于中序遍历过程中,有问题的当前节点node(到目前为止,最后一个遍历到的节点),则为second。

3. 交换二者的val

 

代码参考:

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 8  *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 9  *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10  * };
11  */
12 class Solution {
13 public:
14     TreeNode* first = nullptr, * second = nullptr;
15     TreeNode* pre = nullptr;
16     //inorder traverse the tree,
17     //to find the mistake two node: first & second
18     //    for each current node: judge if (pre->val > node->val) (wrong order)
19     //                                  (pre: update pre = node in each processing end.)
20     //    ->first: the leftest node which is in wrong order.
21     //        (if first has been assigned, in inorder traverse-> already find the leftest node.)
22     //    ->second: the rightest node which is in wrong order.
23     //        (for each process, current node is the rightest node so far.)
24     //swap these two node
25     void recoverTree(TreeNode* root) {
26         traverse(root);
27         swap(first->val, second->val);
28     }
29     void traverse(TreeNode* node) {
30         if(!node) return;
31         //inorder traverse(BST left<root<right)
32         //1.left
33         traverse(node->left);
34         //2.current node
35         if(pre && pre->val > node->val) {
36             if(!first) first = pre;
37             second = node;
38         }
39         pre = node;
40         //3.right
41         traverse(node->right);
42     }
43 };

 

99. Recover Binary Search Tree

标签:NPU   solution   bsp   ima   upd   order   efi   代码   png   

原文地址:https://www.cnblogs.com/habibah-chang/p/14089096.html

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