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:
    vector<TreeLinkNode*>list;
    void dfs(TreeLinkNode* pRootOfTree){//中序遍历
        if(!pRootOfTree)return ;
        dfs(pRootOfTree->left);
        list.push_back(pRootOfTree);
        dfs(pRootOfTree->right);

        
    }

    TreeLinkNode* GetNext(TreeLinkNode* pNode) {
        TreeLinkNode* root=NULL;
        TreeLinkNode* temp=pNode;
        while(temp){//目的找出根节点
            root=temp;
            temp=temp->next;
        }
        dfs(root);
        for(int i=0;i<list.size();i++){
            if(list[i]==pNode&&i+1<list.size()){//找到下一个结点
                return list[i+1];
            }
        }
        return temp;
    }
};