LeetCode-1104. Path In Zigzag Labelled Binary Tree

题目描述：https://leetcode.com/problems/path-in-zigzag-labelled-binary-tree/

解题思路

二叉树的节点全满的话一般都会有些数学规律的，要善于发现数学规律，二叉树层数用n表示（层数从0层开始数），每一层的节点数为2^n （题目表明该树是无限二叉树）；要找的数字用label表示；label 对应所在数组的下标为index表示；这种题我们分析找规律时要同时画树和对应的数组来结合分析。容易发现，当n为偶数时，label = index + 1；当n为奇数时，index = 2^n - 1 + ( 2^(n+1) - 1 - label) = (2^n) * 3 - label - 2 ，label和index的关系确定，剩下的就好办了；对于一棵树，子节点和父节点的关系如图(容易得到父子节点关系：parent = (child - 1) / 2):

搞清节点与节点之间，数组元素与元素间，下标与下标，下标与要找的数之间的关系，相信编码就不是难事了。

算法描述

Step1：先找出label所在的树的层数n和对应所在数组的下标index（其实树和数组都不需要我们真实的构造出来，只是分析解题时借助一下就行），创建存放路径path的list
Step2：判断n的奇偶性，如果是偶数，index+1即是我们所找的节点元素值，把它放入list的头部；如果是奇数，所找的节点元素值为：(2^n) * 3 - index - 2(这是由解题思路中index 和 label之间的关系得来的，此时要注意：这里的label指的是路径path中每个节点的值，不仅仅是题目所要找的label）
Step3：n--更新n的值，往上递推，同时也更新index的值，index = (index - 1)/2
Step4：循环执行Step2和Step3，直到n<0

代码实现-Java

class Solution {public List<Integer> pathInZigZagTree(int label) {int n = 0;while(Math.pow(2,n) < label){if(Math.pow(2,n+1) > label){break;}else{n++;}}int indexOfLabel = 0;if(n % 2 == 0){indexOfLabel = label - 1 ;}else{indexOfLabel = (int) (Math.pow(2,n) * 3 - label - 2);}List<Integer> list = new LinkedList<>();while(n >= 0){if(n % 2 == 0){list.add(0,indexOfLabel + 1);}else{list.add(0,(int) (Math.pow(2,n) * 3 - indexOfLabel - 2));}indexOfLabel = (indexOfLabel - 1)/2;n--;}return list;}
}

运行结果性能数据：

Runtime: 0 ms, faster than 100.00% of Java online submissions for Path In Zigzag Labelled Binary Tree.
Memory Usage: 33.4 MB, less than 100.00% of Java online submissions for Path In Zigzag Labelled Binary Tree.

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LeetCode-1104. Path In Zigzag Labelled Binary Tree

LeetCode-1104. Path In Zigzag Labelled Binary Tree

解题思路

算法描述

代码实现-Java

运行结果性能数据：

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