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28 lines (25 loc) · 933 Bytes
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// 112. Path Sum
// Easy
// 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.
// 思路
// div and conquer.
// 分解：sum 就是路径上各个节点的 val 的和。
// 治：分解到最后，仅有一个节点时，sum 就是 该节点的 val
const hasPathSum = (root, sum) => {
if (!root) return false
if (!root.left & !root.right) return root.val == sum
// hasPathSum(root.left, sum - root.val)
// hasPathSum(root.right, sum - root.left)
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val)
}