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));
}
};
