450. 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:

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

解答如下：

1. 如果待删除节点最多只有一个孩子，孩子节点替换待删除节点，删除节点。

2. 如果待删除节点有两个孩子节点，找到右子树中最小的节点（后继节点），待删除节点的值替换为后继节点值，继续在右子树中删除后继节点。

LeetCode AC代码如下：

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

TreeNode *minNode(TreeNode *root) {
TreeNode *pNode = root;
while (pNode != NULL && pNode->left != NULL) {
pNode = pNode->left;
}
return pNode;
}
};