剑指offer58 二叉树的下一个结点

自己写的

class Solution {
public:
    TreeLinkNode* GetNext(TreeLinkNode* pNode)
    {
        if(pNode == NULL)
            return NULL;
        if(pNode->right != NULL){
            TreeLinkNode* origin = pNode->right;
            while(origin->left != NULL)
                origin = origin->left;
            return origin;
        }
        else{
            TreeLinkNode* father = pNode->next;
            if(father != NULL){
                if(father->left == pNode)
                    return father;
                while(father != NULL && father->right == pNode){
                    pNode = father;
                    father = pNode->next;
                }
            }
            return father;
        }
    }
};

father != NULL && father->right == pNode这个前后顺序不能换，换了就会报以下错误：
　　　　　　　　　　　　段错误:您的程序发生段错误，可能是数组越界，堆栈溢出（比如，递归调用层数太多）等情况引起


书上的写法：

/*
struct TreeLinkNode {
    int val;
    struct TreeLinkNode *left;
    struct TreeLinkNode *right;
    struct TreeLinkNode *next;
    TreeLinkNode(int x) :val(x), left(NULL), right(NULL), next(NULL) {
        
    }
};
*/
class Solution {
public:
    TreeLinkNode* GetNext(TreeLinkNode* pNode)
    {
        if(pNode == NULL)
            return NULL;
        if(pNode->right != NULL){
            TreeLinkNode* next = pNode->right;
            while(next->left != NULL){
                next = next->left;
            }
            return next;
        }
        else{
            TreeLinkNode* nextone = pNode;
            TreeLinkNode* father = pNode->next;
            while(father != NULL && father->left != nextone){
                nextone = father;
                father = nextone->next;
            }
            return father;
        }
    }
};

father != NULL一定要注意，这种情况一个是让father->left能不报错，还有一个就是解决了一个bad case，就是这个结点就是最后一个结点就返回NULL，或者说万一父结点全是右子树的