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* root = pNode;
        while (root->next != NULL)
            root = root->next;
        vector<TreeLinkNode*> vin;

        Inorder(root, vin);

        for (int i = 0; i < vin.size() - 1; i++) {
            if (pNode == vin.at(i)) {
                return vin.at(i+1);
            }
        }
        return NULL;

    }
	// 中序遍历递归
    void Inorder(TreeLinkNode* root, vector<TreeLinkNode*> &vin) {
        if (root == NULL)
            return;
        Inorder(root->left, vin);
        vin.push_back(root);
        Inorder(root->right, vin);
    }
};