今天在leetcode上写了一个关于删除二叉排序树上指定结点的一个题,在待删除结点有左右子树的情况下,找出右子树的最左端的结点;
//第一种;
struct TreeNode* findMin(struct TreeNode *root)
{
struct TreeNode *p;
if(root->left)
p=findMin(root->left);
return root;
}
第二种;
struct TreeNode* Findleftmin(struct TreeNode *root){
if(root->left)
return Findleftmin(root->left);
return root;
}
第一种发生了错误 第二种正确,我没法现问题所在,有大佬能解释一下哪有问题?
完整的代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode* Findleftmin(struct TreeNode *root){
if(root->left)
return Findleftmin(root->left);
return root;
}
struct TreeNode* deleteNode(struct TreeNode* root, int key){
struct TreeNode *p;
if(root==NULL)
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)
{
p=Findleftmin(root->right);
root->val=p->val;
root->right=deleteNode(root->right , root->val);
}
else {
p= root;
if(root->right==NULL && root->left ==NULL) //为叶结点时;
root=NULL;
else if(root->left ) // 仅有左子树;
root=root->left;
else if(root->right) // 仅有右子树;
root=root->right;
free(p);
}
}
return root;
}