112. Path Sum

Easy

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output:
 true
Explanation:
 The root-to-leaf path with the target sum is shown.

Example 2:

Input: root = [1,2,3], targetSum = 5
Output:
 false
Explanation:
 There two root-to-leaf paths in the tree:
(1 --> 2): The sum is 3.
(1 --> 3): The sum is 4.
There is no root-to-leaf path with sum = 5.

Example 3:

Input: root = [], targetSum = 0
Output:
 false
Explanation:
 Since the tree is empty, there are no root-to-leaf paths.

Constraints:

• The number of nodes in the tree is in the range [0, 5000].

• -1000 <= Node.val <= 1000

• -1000 <= targetSum <= 1000

解題

下面是我原本的解法，比較多冗余的程式碼：

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func hasPathSum(root *TreeNode, targetSum int) bool {
    if root==nil  {
        return false
    }
    
    if targetSum-root.Val==0 && root.Left==nil && root.Right==nil {
        return true
    }
    
    if root.Left==nil {
        return hasPathSum(root.Right, targetSum-root.Val) 
    }
    
    if root.Right==nil {
        return hasPathSum(root.Left, targetSum-root.Val) 
    }
    
    leftSide := hasPathSum(root.Left, targetSum-root.Val)
    rightSide := hasPathSum(root.Right, targetSum-root.Val)
    
    if leftSide||rightSide {
        return true
    }
    
    return false
}

下面是改良的精簡版本：

func hasPathSum(root *TreeNode, targetSum int) bool {
    if root==nil { return false }
    
    if root.Left==nil && root.Right==nil { return root.Val == targetSum }
    
    return hasPathSum(root.Left, targetSum-root.Val) || hasPathSum(root.Right, targetSum-root.Val)
}