题目

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.

For 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.

解题思路

判断从根节点到叶子节点的路径节点和是否等于目标值

1. 定义全局变量hasSum用于记录是否有从根节点到叶子节点的路径节点和等于目标值
2. 遍历所有root-to-leaf path，累加得到各路径的值，
3. 如果和等于目标值，则将全局变量hasSum置为true
4. 最后返回hasSum的值

注意
需要判断叶子节点，叶子节点的左右节点都为空

代码

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */

var hasSum bool

func addPathSum(t *TreeNode, sum, sumNow int)  {
    if t == nil {
        return
    } 
    
    val := t.Val
    sumNow += val
    if sum == sumNow && isLeaf(t){
            hasSum = true
    } else {
        addPathSum(t.Left, sum, sumNow)
        addPathSum(t.Right, sum, sumNow)
    }
}

func isLeaf(t *TreeNode) bool {
    if t != nil && t.Left ==nil && t.Right == nil {
        return true
    }   
    return false
}

func hasPathSum(root *TreeNode, sum int) bool {
    hasSum = false
    if root == nil {
        return false
    }

    sumNow := root.Val
    if isLeaf(root) && sum == sumNow {
        hasSum = true
    }
    
    addPathSum(root.Left, sum, sumNow)
    addPathSum(root.Right, sum, sumNow)

        return hasSum
}