[LeetCode] 1430. 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.

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

题意是给一个二叉树和一个从根节点开始的sequence，请你判断这个sequence是否是一个有效的，从根节点到某个叶子节点的sequence。我只放一个例子好了，因为这个题比较直观。如果给出的sequence arr少任何一个节点，都要return false。

思路是DFS深度遍历。如果遍历到的当前节点没有左孩子也没有右孩子，说明是叶子节点了，则判断此时的深度 + 1是不是跟遍历arr的长度一致。需要检查的corner case有

• 当前深度大于arr.length
• 当前节点上的值跟遍历到的arr[index]不一样
• 当前节点为空 - 事实上在路径中的任何一个节点都不能为空

时间O(n)

空间O(n)

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isValidSequence(TreeNode root, int[] arr) {
        return dfs(root, arr, 0);
    }
    
    private boolean dfs(TreeNode n, int[] a, int depth) {
        if (n == null || depth >= a.length || a[depth] != n.val) { // base cases.
            return false;
        }// key base case: a leave found.
        if (n.left == null && n.right == null) { // credit to @The_Legend_ for making the code clean
            return depth + 1 == a.length; // valid sequence?
        }
        return dfs(n.left, a, depth + 1) || dfs(n.right, a, depth + 1); // recurse to the children.
    }
}

LeetCode 题目总结