如何查找二叉搜索树中重复元素的数量
public void duplicate()
{
int repeatation = 0;
Node current = root;
Node duplicate = root;
while (current == null)
{
if (duplicate == current || duplicate == current.right || duplicate== current.left)
{
Console.WriteLine("node is repeated :" + duplicate);
repeatation++;
}
}
Console.WriteLine("number of repeatation is :" + repeatation);
}
这段代码用于二叉搜索树中的重复元素，以及一个元素重复了多少次，但它无法正常工作，你能告诉我这段代码有什么问题吗，我不确定我的代码是否正确......
public void duplicate()
{
int repeatation = 0;
Node current = root;
Node duplicate = root;
while (current == null)
{
if (duplicate == current || duplicate == current.right || duplicate== current.left)
{
Console.WriteLine("node is repeated :" + duplicate);
repeatation++;
}
}
Console.WriteLine("number of repeatation is :" + repeatation);
}
this code is for duplicate elements in binary search tree and how many time a element repeat but its not working properly ,would you tell me whats wrong with this code and i am not sure i code it right or not......
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如果你遍历树 InOrder 你会得到重复的元素在一起，所以你只需要检查一个值何时等于前一个值，以及这种情况发生了多少次。
If you traverse the tree InOrder you'll get the repeated elements together, so you just need to check when a value equals the previous, and how many times this happens.
以下方法可能会帮助您解决此问题：
1）获取 BST 中的最大值（向右，向右，...，向右，直到到达叶子）
2）遵循以下伪代码：
Here's an approach that might help you with this issue:
1) Get the maximum value in your BST (just go right, right, ..., right until you reach a leaf)
2) follow this pseudo code:
您只需遍历 BST 一次，将每个元素保存在哈希映射中，并记录您遇到该元素的次数。
之后，迭代哈希映射并输出计数器大于 1 的每个元素。
时间复杂度：O(n)
You can just traverse your BST once saving every element in a hash-map along with a counter of how many times you have encountered that element.
After that, iterate through your hash-map and output every element that has a counter greater than 1.
Time complexity: O(n)