leetcode 30 days challenge Check If a String Is a Valid Sequence from Root to Leaves Path in a Binary Tree

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].

solution

Depth-first search (DFS) with the parameters: current node in the binary tree and current position in the array of integers.深度优先遍历

When reaching at final position check if it is a leaf node.

class Solution {
public:
    bool isValidSequence(TreeNode* root, vector<int>& arr) {
        return helper(root,arr,0);
    }
    bool helper(TreeNode * node,vector<int>& arr, int pos)
    {
        if(node==NULL)
            return false;
        if(node->val!=arr[pos])
           return false;

        if(pos==arr.size()-1)//数组完了,判断当前节点是不是叶节点
            return (node->left==NULL && node->right==NULL);

        return(helper(node->left,arr,pos+1)||helper(node->right,arr,pos+1));
    }
};

参考链接

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