Delete Node in a BST

Nov 28, 2019

第24天。

今天的题目是 Delete Node in a BST :


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:

  1. Search for a node to remove.
  2. 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

水题,只要先在BST上做搜索,然后删除就好了,因为只是BST,所以可以不考虑平衡的问题:

  • leftright都为空:直接删除,返回nullptr即可
  • leftright都不为空:默认采用把右子树的节点拉上来的方式,即把左子树插入到右子树中,然后再返回right即可。
  • leftright有一个不为空,则返回不为空的子树即可。

则代码如下:

    TreeNode* deleteNode(TreeNode *node) {
        auto left = node->left, right = node->right;
        delete node;
        if (left && right) {
            auto temp = right;
            while(temp->left) {
                temp = temp->left;
            }
            temp->left = left;
            return right;
        } 
        return (left ? left : (right ? right : nullptr));
    }
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (root == nullptr) return nullptr;
        else if (root->val == key) {
            return deleteNode(root);
        } else if (key >  root->val) {
            root->right = deleteNode(root->right, key);    
        } else 
            root->left = deleteNode(root->left, key);
        
        return root;
    }
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