leetcode 450. Delete Node in a BST

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

这个题写了很久！真的最近有点生疏了。

1、涉及删除节点，需要知道删除节点的父节点才好操作。

/**
* 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) return NULL;
TreeNode* q = root;
TreeNode* parent = NULL; //涉及删除节点，需要知道这个删除节点的父节点。

while (q) //定位key值点的位置
{
if (q->val > key)
{
parent = q;
q = q->left;
}
else if (q->val < key)
{
parent = q;
q = q->right;
}
else
{
if (q == root && !root->left)
return root->right;
find_the_point(q, parent); //找到这个节点左子树的最大值，并且替换到当前来
break;
}
}
return root;
}

void find_the_point(TreeNode* root, TreeNode* parent) //找到这个节点左子树的最大值，并且替换到当前来
{
if (!root->left) //如果没有左子树，那就直接删除当前点。
{
TreeNode* del = root;
if (parent && parent->left == root) parent->left = root->right;
else if (parent && parent->right == root) parent->right = root->right;
delete root;
return;
}
parent = root;
TreeNode* lmax = root->left;
while (lmax->right) //找到root左子树的最大值节点。
{
parent = lmax;
lmax = lmax->right;
}

root->val = lmax->val; //把左子树的最大值节点的值换到root上来 当前lmax肯定没有右子树
if (parent && parent->left == lmax) parent->left = lmax->left;
else if (parent && parent->right == lmax) parent->right = lmax->left;
delete lmax;
return;
}
};