• [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

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     int val;
     5  *     TreeNode left;
     6  *     TreeNode right;
     7  *     TreeNode() {}
     8  *     TreeNode(int val) { this.val = val; }
     9  *     TreeNode(int val, TreeNode left, TreeNode right) {
    10  *         this.val = val;
    11  *         this.left = left;
    12  *         this.right = right;
    13  *     }
    14  * }
    15  */
    16 class Solution {
    17     public boolean isValidSequence(TreeNode root, int[] arr) {
    18         return dfs(root, arr, 0);
    19     }
    20     
    21     private boolean dfs(TreeNode n, int[] a, int depth) {
    22         if (n == null || depth >= a.length || a[depth] != n.val) { // base cases.
    23             return false;
    24         }// key base case: a leave found.
    25         if (n.left == null && n.right == null) { // credit to @The_Legend_ for making the code clean
    26             return depth + 1 == a.length; // valid sequence?
    27         }
    28         return dfs(n.left, a, depth + 1) || dfs(n.right, a, depth + 1); // recurse to the children.
    29     }
    30 }

    LeetCode 题目总结

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  • 原文地址:https://www.cnblogs.com/cnoodle/p/12812393.html
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