|
| 1 | +/** |
| 2 | + * Example: |
| 3 | + * var ti = TreeNode(5) |
| 4 | + * var v = ti.`val` |
| 5 | + * Definition for a binary tree node. |
| 6 | + * class TreeNode(var `val`: Int) { |
| 7 | + * var left: TreeNode? = null |
| 8 | + * var right: TreeNode? = null |
| 9 | + * } |
| 10 | + */ |
| 11 | +//https://leetcode.com/problems/path-sum/description |
| 12 | +class Solution { |
| 13 | + fun hasPathSum(root: TreeNode?, targetSum: Int): Boolean { |
| 14 | + return dfs(root, 0, targetSum) |
| 15 | + } |
| 16 | + |
| 17 | + fun dfs(node: TreeNode?, sum: Int, targetSum: Int): Boolean { |
| 18 | + if (node == null) return false |
| 19 | + val currentSum = sum + node.`val` |
| 20 | + if (node.left == null && node.right == null) { |
| 21 | + return currentSum == targetSum |
| 22 | + } |
| 23 | + return dfs(node.left, currentSum, targetSum) || dfs(node.right, currentSum, targetSum) |
| 24 | + } |
| 25 | + |
| 26 | + fun hasPathSum1(root: TreeNode?, targetSum: Int): Boolean { |
| 27 | + if (root == null) return false |
| 28 | + if (root.left == null && root.right == null && root.`val` == targetSum) return true |
| 29 | + return hasPathSum(root.left, targetSum - root.`val`) || hasPathSum(root.right, targetSum - root.`val`) |
| 30 | + } |
| 31 | +} |
| 32 | + |
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