Recover Binary Search Tree

最新推荐文章于 2022-03-13 20:36:15 发布

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最新推荐文章于 2022-03-13 20:36:15 发布

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分类专栏： leetcode习题分析文章标签：二叉搜索树 BST 中序遍历

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本文介绍了一种常数空间解决方案，用于在二叉搜索树中找到并修复由于误操作导致的节点位置错误，通过中序遍历实现定位。

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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?

解法：寻找BST中错位的两个节点其实就是中序遍历BST，找到不符合val递增的两个点。前面分析过中序遍历的三种方法，这里题目要求用O(1)的空间，所以不采用递归和栈的方式。那么通过下面的代码中需遍历BST。

    void InoderTraversal(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function

        TreeNode* cur = root;
        TreeNode* prev = NULL;

        
        while(cur != NULL)
        {
            if(cur->left != NULL)
            {
                prev = cur->left;
                while(prev->right != NULL && prev->right != cur) prev = prev->right;
                if(prev->right == cur){
					//TODO:
                    prev->right = NULL;
                    cur = cur->right;
                }else{
                    prev->right = cur;
                    cur = cur->left;
                }  
            }else{
				//TODO:
                cur  = cur->right;    
            }
        }
    }

在TODO处完成了当前节点的遍历，开始右子树，所以在这里比较val确定错位。

为了定位错位的两个Node，需要一个记录中序遍历的前一个节点的指针preVisited。完整代码如下：

/**
 * 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
        TreeNode* e1 = NULL;
        TreeNode* e2 = NULL;
        TreeNode* cur = root;
        TreeNode* prev = NULL;
        TreeNode* preVisited = NULL;
        
        while(cur != NULL)
        {
            if(cur->left != NULL)
            {
                prev = cur->left;
                while(prev->right != NULL && prev->right != cur) prev = prev->right;                
                if(prev->right == cur){  
                    //TODO:
                    if(preVisited != NULL && preVisited->val > cur->val){
                        if(e1 == NULL){
                            e1 = preVisited;
                        }
                        e2 = cur;  
                    }                    
                    prev->right = NULL;
                    preVisited = cur;
                    cur = cur->right;
                }else{
                    prev->right = cur;
                    cur = cur->left;
                }
                
            }else{
                //TODO:
                if(preVisited != NULL && preVisited->val > cur->val){
                    if(e1 == NULL){
                        e1 = preVisited;
                    }
                    e2 = cur; 
                }            
                preVisited = cur;
                cur  = cur->right;    
            }
        }
        if(e1 != NULL && e2 != NULL) swap(e1->val,e2->val);
    }
};

296 milli secs.