[Swift]LeetCode112. 路径总和 | Path Sum

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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。

---

/**
 * 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 hasPathSum(_ root: TreeNode?, _ sum: Int) -> Bool {
        if root == nil {return false}
        return check(root,0,sum)
    }
    func check(_ node:TreeNode?,_ sum:Int,_ cur:Int)->Bool
    {
        if node == nil && sum == cur
        {
            return true
        }
        if node == nil
        {
            return false
        }
        if node!.left == nil && node!.right != nil 
        {
            return check(node!.right,sum+node!.val,cur)
        }
        if node!.left != nil && node!.right == nil
        {
            return check(node!.left,sum+node!.val,cur)
        }
        return check(node!.left,sum + node!.val, cur)
        ||  check(node!.right,sum + node!.val, cur)
    }
}

---

36ms

/**
 * 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 hasPathSum(_ root: TreeNode?, _ sum: Int) -> Bool {
        if root?.left == nil, root?.right == nil { // 叶节点或空节点
            guard root != nil else {
                return false
            }
            return sum == root!.val ? true : false
        }
        return hasPathSum(root?.left, sum - root!.val) || hasPathSum(root?.right, sum - root!.val)
    }
}

---

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 hasPathSum(_ root: TreeNode?, _ sum: Int) -> Bool {
        
        guard let root = root else { return false }
        
        let difference = sum - root.val
        if difference == 0 && root.left == nil && root.right == nil {
            return true
        } else {
            return hasPathSum(root.left, difference) || hasPathSum(root.right, difference)
        }
        
        return false
    }
}