Algorithm
Problem Name: 1104. Path In Zigzag Labelled Binary Tree
In an infinite binary tree where every node has two children, the nodes are labelled in row order.
In the odd numbered rows (ie., the first, third, fifth,...), the labelling is left to right, while in the even numbered rows (second, fourth, sixth,...), the labelling is right to left.
Given the label of a node in this tree, return the labels in the path from the root of the tree to the node with that label.
Example 1:
Input: label = 14 Output: [1,3,4,14]
Example 2:
Input: label = 26 Output: [1,2,6,10,26]
Constraints:
1 <= label <= 10^6
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const pathInZigZagTree = function(label) {
const res = [], { log2, floor, ceil } = Math
const level = floor(log2(label))
let compl = 2 ** (level + 1) - 1 + 2 ** level - label
while(label) {
res.push(label)
label = floor(label / 2)
compl = floor(compl / 2)
;[label, compl] = [compl, label]
}
res.reverse()
return res
};
Input
Output
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def pathInZigZagTree(self, label: int) -> List[int]:
res = [label]
while label != 1:
d = int(math.log(label) / math.log(2))
offset = int(2 ** (d + 1)) - label - 1
label = (int(2 ** d) + offset) // 2
res = [label] + res
return res
Input
Output
#3 Code Example with C# Programming
Code -
C# Programming
using System.Collections.Generic;
namespace LeetCode
{
public class _1104_PathInZigzagLabelledBinaryTree
{
public IList < int> PathInZigZagTree(int label)
{
int nodeCount = 1;
while (label >= nodeCount * 2)
nodeCount *= 2;
var result = new List < int>();
while (label > 0)
{
result.Add(label);
var maxLevellable = nodeCount * 2 - 1;
var minLevellable = nodeCount;
var parent = (maxLevellable + minLevellable - label) / 2;
label = parent;
nodeCount /= 2;
}
result.Reverse();
return result;
}
}
}
Input
Output