题解 | 二叉树的下一个结点

/*
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) {
        TreeLinkNode* rightNode = pNode->right;
        if(rightNode!=nullptr){
            while(rightNode->left!=nullptr){
                rightNode = rightNode->left;
            }
            return rightNode;
        }
        else{
            TreeLinkNode* res = pNode;
            while(res->next){
                if(res->next->left == res) return res->next;
                res = res->next;
            }
        }
        return nullptr;
        
    }
};