### 19885911436 # 兰州高端桑拿场子_每每想起那触感都想再来一次的冲动

2021-10-22 13:45:04 浏览次数：

If we want to del兰州高端桑拿场子_每每想起那触感都想再来一次的冲动

【兰州桑拿休闲中心】-【兰州洗浴水疗会所】-【兰州顶级水疗会所】-【兰州品茶休闲会所】-【兰州休闲洗浴中心】-【兰州精英男士水疗会所】ete 15 from the above BST, we can do some tricks to reduce the situation to either case 1 or case 2.

#### FIND THE MINIMAL VALUE OF ITS RIGHT SUBTREE If we find the minimal value of its right subtree, it should not be node with two children, otherwise the node’s left child will be smaller than 1. Given the above BST, the minimal value of 15’s subtree will be 17.

Then we replace the to-be-deleted value with 17, we then have two 17’s. So the next task is to delete the 17 from the original 15’s right sub tree. So this is the case of deleting node with only 1 children.

Why this works? Because the 17 is on the 15’s right subtree, so it should be greater than 15, which is also greater than any other nodes in the 15’s left subtree. 17 is also the minimal value in the 15’s right subtree, so 17 is less or equal than any of 15’s right sub tree. Of course we have a duplicate 17 after replacing 15 with the value 17.

#### FIND THE MAXIMUM VALUE OF ITS LEFT SUBTREE

Similarly, we can find the maximum value of the to-be-deleted node’s left subtree and the proof/approach is similar. Another to note is that if the value found (either maximum or minimal) has no children, then we are reducing the case to case 1 where deleting a leaf node from a BST tree.

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