LeetCode #450 - Delete Node in a BST

最新推荐文章于 2021-06-09 15:43:37 发布

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本文介绍了一种在二叉搜索树(BST)中删除指定键值节点的算法。通过递归查找目标节点，根据节点的子树情况，直接替换或调整树结构，确保删除操作后树的BST性质不变。提供了C++实现代码示例。

题目描述：

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

/ \

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.

/ \

4 6

/ \

2 7

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

/ \

2 6

\ \

4 7

利用递归找到要删除的节点，如果该节点只有左子树或者右子树，那么直接让该节点等于左子树或者右子树，否则就要找到右子树的最小值，将它换到当前节点，然后再在右子树中删除最小值节点。

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if(!root) return NULL;
        if(root->val>key) root->left=deleteNode(root->left,key);
        else if(root->val<key) root->right=deleteNode(root->right,key);
        else 
        {
            if (!root->left||!root->right) 
                root=(root->left)?root->left:root->right;
            else 
            {
                TreeNode *cur=root->right;
                while(cur->left) cur=cur->left;
                root->val=cur->val;
                root->right=deleteNode(root->right,cur->val);
            }
        }
        return root;
    }
};