[Swift]LeetCode113. 路径总和 II | Path Sum II

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Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \    / \
7    2  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]

---

给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

说明: 叶子节点是指没有子节点的节点。

示例:
给定如下二叉树，以及目标和 sum = 22，

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

返回:

[
   [5,4,11,2],
   [5,8,4,5]
]

---

24ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func pathSum(_ root: TreeNode?, _ sum: Int) -> [[Int]] {
        var resArray =  [[Int]]()
        if root == nil {
            return resArray
        }

        if root?.left == nil && root?.right == nil {
            if root!.val == sum {
                let nodeArray = [Int](repeating:root!.val,count:1)
                resArray.append(nodeArray)
            }
            return resArray
        }

        var nodeArray =  [Int]()
        helper(root,sum,&resArray,&nodeArray)
        return resArray
    }

    func helper(_ root: TreeNode?, _ sum: Int, _ array: inout [[Int]], _ nodeArray: inout [Int]) {
        if root == nil {
            return
        }

        if root?.left == nil && root?.right == nil {
            if sum-root!.val == 0 {
                nodeArray.append(root!.val)
                array.append(nodeArray)
                nodeArray.remove(at:nodeArray.count-1)
            }
            return
        }
        
        nodeArray.append(root!.val)
        helper(root?.left,sum-root!.val,&array,&nodeArray) 
        helper(root?.right,sum-root!.val,&array,&nodeArray)
        let index = nodeArray.count - 1
        nodeArray.remove(at:index)
    }
}

---

28ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
*/
class Solution {
    func pathSum(_ root: TreeNode?, _ sum: Int) -> [[Int]] {
        var ans = [[Int]]()
        var combination = [Int]()
        pathSum(root, sum, &combination, &ans)
        return ans
    }
    
    func pathSum(_ node: TreeNode?, _ sum: Int, _ combination: inout [Int], _ ans: inout [[Int]]) {
        guard let node = node else { return }
        
        combination.append(node.val)
        
        if node.left == nil && node.right == nil && (sum - node.val) == 0 {
            ans.append(combination)
        }     
        
        pathSum(node.left, sum - node.val, &combination, &ans)
        pathSum(node.right, sum - node.val, &combination, &ans)
        
        combination.removeLast()
    }
}

---

32ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
        func pathSum(_ root: TreeNode?, _ sum: Int) -> [[Int]] {
        if root == nil { return []}
        var result = [[Int]]()
        var currentRun = [Int]()
        checkSum(root, sum, &result, &currentRun)
        return result
        
    }
    func checkSum(_ root: TreeNode?, _ sum: Int,_ result: inout [[Int]],_ currentRun: inout [Int] ) {
        if root == nil { return }
        var current = currentRun
        current.append(root!.val)
        if sum - root!.val == 0 && root!.left == nil && root!.right == nil {
            result.append(current)
            return
        }
        checkSum(root!.left, sum - root!.val, &result, &current)
        checkSum(root!.right, sum - root!.val, &result, &current)
    }
}

---

52ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func pathSum(_ root: TreeNode?, _ sum: Int) -> [[Int]] {
        var paths: [[Int]] = []
        BFS(root, sum: sum, paths: &paths)
        return paths
    }
    
    func BFS(_ root: TreeNode?, sum: Int, paths: inout [[Int]], path: [Int]? = nil) {
        guard let root = root else { return }
        var path: [Int] = path ?? []
        path.append(root.val)
        if root.left == nil && root.right == nil {
            if sum == path.reduce(0) { $0 + $1 } {
                paths.append(path)
            }
            return
        }
        BFS(root.left, sum: sum, paths: &paths, path: path)
        BFS(root.right, sum: sum, paths: &paths, path: path)
    }
}

---

 56ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func pathSum(_ root: TreeNode?, _ sum: Int) -> [[Int]] {
        guard let root = root else {
            return []
        }
    
        if root.left == nil && root.right == nil && root.val == sum {
            return [[sum]]
        }
    
        let lPathSum = pathSum(root.left, sum - root.val)
        let rPathSum = pathSum(root.right, sum - root.val)
    
        return (lPathSum + rPathSum).map {
            [root.val] + $0
        }
    }
}