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[Golang][Leetcode][BinaryTree]刷題系列-112-Path Sum


112. Path Sum


Level : Easy

原題連結 : Click

題目 :

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 :

Note

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 Output: true

Example 2:

Input: root = [1,2,3], targetSum = 5 Output: false

Example 3:

Input: root = [1,2], targetSum = 0 Output: false


解題思路 :

  • 這題需要我把檢查是否有任何一個path的和等於他給的targetSum
  • 這題考察的其實的不外乎就是對dfs的概念,加上若是用recursive的方式會更簡潔,所以一方面也考察對於recursive code的掌握度,誰作為回傳值or誰作為argument…等
  • 解法1是我一開始的答案,非常直觀的寫法,且time complexity是O(n)
  • 看完解法1可以繼續看我從討論區得到的解法2,這個解法就更簡潔了,是利用targetSum進行減法去做檢查,是不是更高明了呢!? 趕快學起來!

Recursive解法-1

  • time complexity: O(n) , space complexity: O(1)

Runtime: 4 ms, faster than 91.44% of Go online submissions for Path Sum.


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/**
 * 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
    }
    
    res := false
    var sum int
    
    
    findTargetSum(root, &res, sum, targetSum)
    
    return res
    
}

func findTargetSum(node *TreeNode, res *bool, sum int, targetSum int) {
     
    if *res == true {
        return 
    }

    sum += node.Val
    
    
    if node.Left == nil && node.Right == nil {
        
        if targetSum == sum {
            *res = true
        }
    }
    
    if node.Left != nil {
        findTargetSum(node.Left,res,sum, targetSum)
    }
    
    if node.Right != nil {
        findTargetSum(node.Right,res,sum, targetSum)
    }
    
    return 
}

Recursive解法-2

  • time complexity: O(n) , space complexity: O(1)

Runtime: 4 ms, faster than 91.44% of Go online submissions for Path Sum.


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/**
 * 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 root.Left == nil && root.Right == nil {
        return targetSum == root.Val
    }
    return hasPathSum(root.Left, targetSum - root.Val) || hasPathSum(root.Right, targetSum - root.Val)
}


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