[Swift]LeetCode1104. 二叉树寻路 | Path In Zigzag Labelled Binary Tree

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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

---

在一棵无限的二叉树上，每个节点都有两个子节点，树中的节点 逐行 依次按 “之” 字形进行标记。

如下图所示，在奇数行（即，第一行、第三行、第五行……）中，按从左到右的顺序进行标记；

而偶数行（即，第二行、第四行、第六行……）中，按从右到左的顺序进行标记。

给你树上某一个节点的标号 label，请你返回从根节点到该标号为 label 节点的路径，该路径是由途经的节点标号所组成的。

示例 1：

输入：label = 14
输出：[1,3,4,14]

示例 2：

输入：label = 26
输出：[1,2,6,10,26]

提示：

• 1 <= label <= 10^6

---

Runtime: 4 ms

Memory Usage: 20.5 MB

class Solution {
    func pathInZigZagTree(_ label: Int) -> [Int] {
        var ret:[Int] = [Int]()
        var cur:Int = label
        while(true)
        {
            ret.insert(cur,at:0)
            if cur == 1 {break}
            cur /= 2
            let h:Int = highestOneBit(cur)
            cur ^= h-1
        }
        return ret
    }
    
    func highestOneBit(_ i:Int) -> Int
    {
        var i = i
        // HD, Figure 3-1
        i |= (i >>  1)
        i |= (i >>  2)
        i |= (i >>  4)
        i |= (i >>  8)
        i |= (i >> 16)
        return i - (i >> 1)
    }
}

---

8ms 

class Solution {
    func pathInZigZagTree(_ label: Int) -> [Int] {
        var result = [Int]()
        var log = 1
        while log <= label {
            log <<= 1
        }
        var label = label 
        while label >= 1 {
            result.append(label)
            label = log + log / 2 - label - 1
            label /= 2
            log /= 2
        }
        return result.reversed()
    }
}

---

316ms

class Solution {
    func pathInZigZagTree(_ label: Int) -> [Int] {
        var level = 1 
        var arr = [Int]()
        
        outer: while true {
            let left = Int(pow(2.0, Double(level-1)))
            let right = Int(pow(2.0, Double(level))) - 1

            if level % 2 == 1 {
                for v in left...right {
                    if v == label { 
                        arr.append(v)
                        break outer
                    }
                    arr.append(v)
                }
            } else {
                for v in stride(from:right, through: left, by: -1) {
                    if v == label { 
                        arr.append(v)
                        break outer
                    }
                    arr.append(v)
                }
            }
            level += 1
        }
        
        var idx = arr.count - 1
        var result = [Int]()
        
        while idx > 0 {
            result.append(arr[idx])
            idx = (idx - 1) / 2
        }
        result.append(arr[0])
        return result.reversed()
    }
}

---

536ms

class Solution {
    func pathInZigZagTree(_ label: Int) -> [Int] {
        var tree = [Int]()
        var prevPow = 1
        var reverse = false
        outer: for i in 1...20 {
            let num = Int(pow(2, Double(i)))
            if reverse {
                for j in stride(from: num - 1, through: prevPow, by: -1) {
                    tree.append(j)
                    if j == label { break outer }
                }
            } else {
                for j in stride(from: prevPow, to: num, by: 1) {
                    tree.append(j)
                    if j == label { break outer }
                }
            }
            prevPow = num
            reverse = !reverse
        }
        var res = [Int]()
        var i = tree.count - 1
        while i > 0 {
            res.append(tree[i])
            i = (i - 1)/2
        }
        res.append(tree[i])
        return Array(res.reversed())
    }
}