算法第七周Delete Node in a BST[medium]

Delete Node in a BST[medium]

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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

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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：

1. 
TreeNode* temp = left;
while (temp->right != NULL) {
temp = temp->right;
}
temp->right = right->left;
right->left = left;

2. 
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;        }        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;    }};