LeetCode详细分析 :: Recover Binary Search Tree [Tree]


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?

confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.

这里有几个知识点可以注意一下,是个不错的考察BST的题目。一、要如何找出这两个有乱序的位置呢?如果直接在树上考察,似乎关系有些模糊;那么,考虑一个性质,通过中序遍历,可以将BST按升序输出,eg:12
3 4 56 7,这里任意调换一下位置,构成16 3 4 5 2 7, 6 和
2将数列分割成了三份,每份独立时仍然保持升序(有可以能头尾的部门不包含元素),但是6 > 3, 5 > 2;说明,第一次出现 pre > cur的情况,pre是那个错误节点,第二次出现 pre > cur的情况,cur是错误节点,记录下来就可以了。

二、有可能调换位置的元素在排序中相邻(树结构图中不一定相邻),一中提到的pre >cur的情况就发生重合,只有一次;

代码如下:

class Solution {
private:
TreeNode *Pre, *node1, *node2;  //申明为成员变量,这样每个函数都可以访问,不必重复创建;

void inOrder(TreeNode *root){
    if (root == NULL)
        return;
    inOrder(root->left);
    if (Pre != NULL && Pre->val > root->val){  //第二次(也是最后一次)出现的才是正确位置,这里这样写避免了判断,直接覆盖
        node2 = root;
        if (node1 == NULL)             //利用 NULL 为是否为第一次出现的标记
            node1 = Pre;
    }
    Pre = root;
    inOrder(root->right);
}

public:
    void recoverTree(TreeNode *root) {
        Pre = NULL;
        node2 = node1 = NULL;

        inOrder(root);

        node1->val ^= node2->val;
        node2->val ^= node1->val;
        node1->val ^= node2->val;
    }
};

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时间: 2024-10-17 17:17:28

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