leetcode-103-binary-tree-zigzag-level-order-traversal.swift

/**

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https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/


Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree [3,9,20,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its zigzag level order traversal as:
[
  [3],
  [20,9],
  [15,7]
]

*/

/**
Big O Annotation
Time complexity O(n) where n is the amount of nodes in the BST.
Space complexity O(n) where n is the amount of nodes in the BST.
*/
func zigzagLevelOrder(_ root: TreeNode?) -> [[Int]] {
        
    // If we have no root, return an empty array
    guard let root = root else {
        return []
    }
    
    // Init a queue...
    var queue = Queue()

    /**
    ... and append the root node and the level 0
    Level 0 because it's a 0 based index top down.
    */
    queue.enqueue(root, atLevel: 0)
    
    /**
    Init a temporary dictionary 
    of level and node values
    */
    var levels: [Int: [Int]] = [:]
    
    // While we still have nodes in the queue...
    while !queue.isEmpty {
        
        // Get the top most node/level tuple
        let (node, level) = queue.dequeue()!
        
        // Check if the level already exists in the tree
        if let mapped = levels[level] {

            // If so, add the value to the array (by combining)
            levels[level] = mapped + [node.val]
        } else {

            // Otherwise add the value as array
            levels[level] = [node.val]
        }
        
        /**
        Add the left node to the queue 
        and increment the level
        */
        if let left = node.left {
            queue.enqueue(left, atLevel: level + 1)
        }
        
        // Same for the right node
        if let right = node.right {
            queue.enqueue(right, atLevel: level + 1)
        }
    }
    
    // Init the output array
    var resultArray: [[Int]] = Array(repeating: [], count: levels.count)
    
    // Iterate through the level dictionary
    for (key, value) in levels {
        
        /**
        Check if the current level is odd
        to get the zig zack effect by reversing the values
        */
        let isOdd = (key % 2) != 0
        resultArray[key] = isOdd ? value.reversed() : value
    }
    
    // And done 🙂
    return resultArray
}

/**
QueueStack implementation with 2 private stacks.
One for enqueueing and the other for dequeueing.

Time complexity O(1)
This API gives us contant time for enqueue and dequeue.
*/
struct Queue {
    private var enqueueStack: [(node: TreeNode, level: Int)] = []
    private var dequeueStack: [(node: TreeNode, level: Int)] = [] 
    
    init() {}
    
    public var isEmpty: Bool {
        return enqueueStack.isEmpty && dequeueStack.isEmpty
    }
    
    public mutating func enqueue(_ node: TreeNode, atLevel level: Int) {
        enqueueStack.append((node: node, level: level))
    }
    
    public mutating func dequeue() -> (node: TreeNode, level: Int)? {
        if dequeueStack.isEmpty {
            dequeueStack = enqueueStack.reversed()
            enqueueStack.removeAll()
        }
        
        return dequeueStack.popLast()
    }
}