[Swift]LeetCode112.路径总和|PathSum

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Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path 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      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，这条路径上所有节点值相加等于目标和。

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      1

返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func hasPathSum(_ root: TreeNode?, _ sum: Int) -> Bool {
16         if root == nil {return false}
17         return check(root,0,sum)
18     }
19     func check(_ node:TreeNode?,_ sum:Int,_ cur:Int)->Bool
20     {
21         if node == nil && sum == cur
22         {
23             return true
24         }
25         if node == nil
26         {
27             return false
28         }
29         if node!.left == nil && node!.right != nil 
30         {
31             return check(node!.right,sum+node!.val,cur)
32         }
33         if node!.left != nil && node!.right == nil
34         {
35             return check(node!.left,sum+node!.val,cur)
36         }
37         return check(node!.left,sum + node!.val, cur)
38         ||  check(node!.right,sum + node!.val, cur)
39     }
40 }

36ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func hasPathSum(_ root: TreeNode?, _ sum: Int) -> Bool {
16         if root?.left == nil, root?.right == nil { // 叶节点或空节点
17             guard root != nil else {
18                 return false
19             }
20             return sum == root!.val ? true : false
21         }
22         return hasPathSum(root?.left, sum - root!.val) || hasPathSum(root?.right, sum - root!.val)
23     }
24 }

28ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func hasPathSum(_ root: TreeNode?, _ sum: Int) -> Bool {
16         
17         guard let root = root else { return false }
18         
19         let difference = sum - root.val
20         if difference == 0 && root.left == nil && root.right == nil {
21             return true
22         } else {
23             return hasPathSum(root.left, difference) || hasPathSum(root.right, difference)
24         }
25         
26         return false
27     }
28 }