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).
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解析
这道题目重点分以下几种情况去分析
- 如果要删除的那个节点存在，且他的左右子树不都存在。那就直接用子树去替代被删除的节点。
- 如果要删除的节点左右子树都存在的话，正如上面所示的其实有两种解决方案，一种就是拿右子树的最小值去替代，另外一种就是拿左子树的最大值去替代。替代完之后又要删除替代的那个节点（这里逻辑就可以套用1去解决）。
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40/**
* 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;
//搜索那个要删除的节点
if(key < root->val){
//在左子树搜
root->left = deleteNode(root->left,key);
} else if(key > root->val){
//在右子树搜
root->right = deleteNode(root->right,key);
} else {
//找到删除节点，而且左右子树只存在一个以下。
if(!root->left||!root->right){
root = (root->left) ? root->left : root->right;
}else{
//左右子树都存在
TreeNode* cur;
cur = root->right;
while(cur->left){
cur = cur->left ;
}
root->val = cur->val;
//删除右子树下用来替换的节点。
root->right = deleteNode(root->right,cur->val);
}
}
return root;
}
};
