算法第七周Delete Node in a BST[medium]

Delete Node in a BST[medium]

Description

Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.

Basically, the deletion can be divided into two stages:

Search for a node to remove.

If the node is found, delete the node.

Note: Time complexity should be O(height of tree).

Example:

root = [5,3,6,2,4,null,7]

key = 3

5
/ \
3   6
/ \   \
2  4   7

Given key to delete is 3. So we find the node with value 3 and delete it.

One valid answer is [5,4,6,2,null,null,7], shown in the following BST.

5
/ \
4   6
/     \
2       7

Another valid answer is [5,2,6,null,4,null,7].

5
/ \
2   6
\   \
4   7

Solution

这道题目是删除BST中的节点。首先我们要对于二叉搜索树有基本的了解，BST一个最基本的特点就是任意一个节点，它的右子树一定是大于它的数，左子数一定是小于它的数，可以利用这一特点进行搜索以及删除。

这一题，我仍然采用了递归的方法。

首先比较root->val的值与key的值。

root->val > key  将key与root的左子树继续比较
root->val < key  将key与root的右子树继续比较
root->val == key  删除结点

在删除节点的时候可以采取两种方式：

TreeNode* temp = left;

while (temp->right != NULL) {

temp = temp->right;

}

temp->right = right->left;

right->left = left;

TreeNode* temp = right;

while (temp->left != NULL) {

temp = temp->left;

}

temp->left = left->right;

left->right = right;

完整代码：

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* deleteNode(TreeNode* root, int key) {
if (root == NULL) return root;
if (root->val > key) {
root->left = deleteNode(root->left, key);
return root;

4000
}
else if (root->val < key) {
root->right = deleteNode(root->right, key);
return root;
}
TreeNode* left = root->left;
TreeNode* right = root->right;
delete root;
if (left == NULL) return right;
if (right == NULL) return left;
TreeNode* temp = right;
while (temp->left != NULL) {
temp = temp->left;
}
temp->left = left->right;
left->right = right;
return left;
}
};