1367. Linked List in Binary Tree

https://leetcode.com/problems/linked-list-in-binary-tree/

Given a binary tree root and a linked list with head as the first node.

Return True if all the elements in the linked list starting from the head correspond to some downward path connected in the binary tree otherwise return False.

In this context downward path means a path that starts at some node and goes downwards.

Example 1:

Input: head = [4,2,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
Output: true
Explanation: Nodes in blue form a subpath in the binary Tree.  

Example 2:

Input: head = [1,4,2,6], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
Output: true

Example 3:

Input: head = [1,4,2,6,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
Output: false
Explanation: There is no path in the binary tree that contains all the elements of the linked list from head.

Constraints:

• 1 <= node.val <= 100 for each node in the linked list and binary tree.

• The given linked list will contain between 1 and 100 nodes.

• The given binary tree will contain between 1 and 2500 nodes.

给定linked list和二叉树，问二叉树中是否存在向下的path和list的值相等。类似字符串匹配的问题，可以主函数做分治并用另一个dfs函数对当前子树做list匹配，时间复杂度O(N * M)。注意另一个用于匹配的dfs不能少，写成以下是错的：

class Solution {
public:
    bool isSubPath(ListNode* head, TreeNode* root) {
        if (head == nullptr) return true;
        if (root == nullptr) return false;
        if (head->val == root->val) {         
            if (isSubPath(head->next, root->left) || isSubPath(head->next, root->right)) return true;
        }
        return isSubPath(head, root->left) || isSubPath(head, root->right);
    }
};

问题在与当head->val != root->val时匹配没有及时返回，如l = [1,2,6] t = [1,4,2,6]时当4和2不等时依然会继续往下分治，而后面是能匹配上的因此整体会错误的返回true。

做串匹配还可以KMP，KMP本质可看作是dp，dp[i]存以pattern字符串A中以i为底可匹配的最长前后缀长度。ptr初始为dp[i -1]，当A[i]和dp[ptr]不等时，ptr = dp[ptr]，直到相等或ptr == 0。比如aacaad的dp = [0,1,0,1,2,0]，匹配d时先检查d和c是否相等，对应aac和aad；不等时往前跳到下一个与A[i-1]匹配的位置，匹配a和d，对应aa和ad。过程如下，方块为未知i，△为prefix中下一个和A[i]匹配的字符：

生成dp数组后遍历树并做匹配，当不满足后把当前已匹配的看作suffix，把待匹配往前移动到对应prefix位置。时间复杂度O(N)。

思路参考自lee251。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSubPath(ListNode *head, TreeNode *root) {
        vector<int> A = {head->val}, dp = {0};
        int ptr = 0;
        head = head->next;
        while (head) {
            while (head && ptr && head->val != A[ptr]) ptr = dp[ptr - 1];
            ptr += head->val == A[ptr];
            A.push_back(head->val);
            dp.push_back(ptr);
            head = head->next;
        }
        return dfs(root, 0, A, dp);
    }

    bool dfs(TreeNode* root, int i, vector<int> &A, vector<int> &dp) {
        if (!root) return false;
        while (i && root->val != A[i]) i = dp[i - 1];
        i += root->val == A[i];
        return i == dp.size() || dfs(root->left, i, A, dp) || dfs(root->right, i, A, dp);
    }
};