回溯法求解Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree

Leetcode三十天挑战的最后一道题，果然越往后leetcode越加大难度，先是出Hard级别的题目，又是出新的题目。果然对我这种小白是够挑战的。还好今天的题我通过回溯法解决了，在官方的提示中，它提示是采用DFS，但是如果采用DFS我就需要额外为每个点建立标记，所以想了想还是用回溯法，风险就是，有可能超时，但好处是答案清晰明了。

Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree
Given a binary tree where each path going from the root to any leaf form a valid sequence, check if a given string is a valid sequence in such binary tree. We get the given string from the concatenation of an array of integers arr and the concatenation of all values of the nodes along a path results in a sequence in the given binary tree.

Example 1:

Input: root = [0,1,0,0,1,0,null,null,1,0,0], arr = [0,1,0,1]

Output: true

Explanation:

The path 0 -> 1 -> 0 -> 1 is a valid sequence (green color in the figure).

Other valid sequences are:

0 -> 1 -> 1 -> 0

0 -> 0 -> 0

Example 2:

Input: root = [0,1,0,0,1,0,null,null,1,0,0], arr = [0,0,1]

Output: false

Explanation: The path 0 -> 0 -> 1 does not exist, therefore it is not even a sequence.

Example 3:

Input: root = [0,1,0,0,1,0,null,null,1,0,0], arr = [0,1,1]

Output: false

Explanation: The path 0 -> 1 -> 1 is a sequence, but it is not a valid sequence.

Constraints:

• 1 <= arr.length <= 5000
• 0 <= arr[i] <= 9
• Each node’s value is between [0 - 9].

Hint1 Depth-first search (DFS) with the parameters: current node in the binary tree and current position in the array of integers.
Hint2 When reaching at final position check if it is a leaf node.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isValidSequence(TreeNode* root, vector<int>& arr) {
        int size = arr.size();
        if(root == NULL && size == 0)
            return true;
        if(root == NULL)
            return false;
        if(size == 0)
            return false;
        if(root->val != arr[0])
            return false;
        vector<int>next(arr.begin() + 1, arr.end());
        if(next.size() == 0)
        {
            if(root->left == NULL & root->right == NULL)
                return true;
            else
                return false;
        }
        if(isValidSequence(root->left, next))
            return true;
        if(isValidSequence(root->right, next))
            return true;
        return false;
    }
};