LeetCode #1104 — MEDIUM

Path In Zigzag Labelled Binary Tree

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Tree

Example 1

Example 2

Core Insight

What unlocks the optimal approach

Based on the label of the current node, find what the label must be for the parent of that node.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1104: Path In Zigzag Labelled Binary Tree
class Solution {
    public List<Integer> pathInZigZagTree(int label) {
        int x = 1, i = 1;
        while ((x << 1) <= label) {
            x <<= 1;
            ++i;
        }
        List<Integer> ans = new ArrayList<>();
        for (; i > 0; --i) {
            ans.add(label);
            label = ((1 << (i - 1)) + (1 << i) - 1 - label) >> 1;
        }
        Collections.reverse(ans);
        return ans;
    }
}

// Accepted solution for LeetCode #1104: Path In Zigzag Labelled Binary Tree
func pathInZigZagTree(label int) (ans []int) {
	x, i := 1, 1
	for x<<1 <= label {
		x <<= 1
		i++
	}
	for ; i > 0; i-- {
		ans = append(ans, label)
		label = ((1 << (i - 1)) + (1 << i) - 1 - label) >> 1
	}
	for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
		ans[i], ans[j] = ans[j], ans[i]
	}
	return
}

# Accepted solution for LeetCode #1104: Path In Zigzag Labelled Binary Tree
class Solution:
    def pathInZigZagTree(self, label: int) -> List[int]:
        x = i = 1
        while (x << 1) <= label:
            x <<= 1
            i += 1
        ans = [0] * i
        while i:
            ans[i - 1] = label
            label = ((1 << (i - 1)) + (1 << i) - 1 - label) >> 1
            i -= 1
        return ans

// Accepted solution for LeetCode #1104: Path In Zigzag Labelled Binary Tree
struct Solution;

impl Solution {
    fn path_in_zig_zag_tree(mut label: i32) -> Vec<i32> {
        let mut level = 0;
        while label >= 1 << level {
            level += 1;
        }
        let mut res = vec![0; level];
        for i in (0..level).rev() {
            res[i] = label;
            let r = (1 << i) - 1;
            let l = r / 2 + 1;
            label = l + r - label / 2;
        }
        res
    }
}

#[test]
fn test() {
    let label = 14;
    let res = vec![1, 3, 4, 14];
    assert_eq!(Solution::path_in_zig_zag_tree(label), res);
    let label = 26;
    let res = vec![1, 2, 6, 10, 26];
    assert_eq!(Solution::path_in_zig_zag_tree(label), res);
    let label = 16;
    let res = vec![1, 3, 4, 15, 16];
    assert_eq!(Solution::path_in_zig_zag_tree(label), res);
}

// Accepted solution for LeetCode #1104: Path In Zigzag Labelled Binary Tree
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1104: Path In Zigzag Labelled Binary Tree
// class Solution {
//     public List<Integer> pathInZigZagTree(int label) {
//         int x = 1, i = 1;
//         while ((x << 1) <= label) {
//             x <<= 1;
//             ++i;
//         }
//         List<Integer> ans = new ArrayList<>();
//         for (; i > 0; --i) {
//             ans.add(label);
//             label = ((1 << (i - 1)) + (1 << i) - 1 - label) >> 1;
//         }
//         Collections.reverse(ans);
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(h)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.