题解 | #二叉树的下一个结点#
二叉树的下一个结点
http://www.nowcoder.com/practice/9023a0c988684a53960365b889ceaf5e
typedef struct TreeLinkNode BNode;
BNode* find_In(BNode* root)
{
BNode* tmp;
if (root == NULL)
return NULL;
tmp = root;
while (!tmp&&!(tmp->left))
{
tmp = tmp->left;
}
return tmp->left;
}
BNode* Find_Next2( BNode* p)
{
if (p->next == NULL)
return NULL;
else {
if (p->next->left == p)
return p->next;
else
return Find_Next2(p->next);
}
}
BNode* Find_Next(BNode* p)
{
if (!p)
return NULL;
if (p->right != NULL && p->right->left != NULL)
return (find_In(p->right));
else if (p->right != NULL)
{
return (p->right);
}
else if (p->next == NULL)
return NULL;
else {
if (p->next->left == p)
return p->next;
else
return Find_Next2(p->next);
}
}
struct TreeLinkNode* GetNext(struct TreeLinkNode* pNode ) {
BNode* next;
if(pNode==NULL)
return NULL;
next=Find_Next(pNode);
return next;
}
