|
| 1 | +/* |
| 2 | +# Time Complexity: O(n) |
| 3 | +# Space Complexity: O(n) |
| 4 | +
|
| 5 | +heap을 사용한 풀이, in-order traversal을 사용한 풀이 풀어볼 것! |
| 6 | +*/ |
| 7 | +/** |
| 8 | + * Definition for a binary tree node. |
| 9 | + * public class TreeNode { |
| 10 | + * int val; |
| 11 | + * TreeNode left; |
| 12 | + * TreeNode right; |
| 13 | + * TreeNode() {} |
| 14 | + * TreeNode(int val) { this.val = val; } |
| 15 | + * TreeNode(int val, TreeNode left, TreeNode right) { |
| 16 | + * this.val = val; |
| 17 | + * this.left = left; |
| 18 | + * this.right = right; |
| 19 | + * } |
| 20 | + * } |
| 21 | + */ |
| 22 | +class Solution { |
| 23 | + public int kthSmallest(TreeNode root, int k) { |
| 24 | + Map<TreeNode, Integer> sizes = new HashMap<>(); |
| 25 | + |
| 26 | + calculateSize(root, sizes); |
| 27 | + |
| 28 | + return findKth(root, k, sizes); |
| 29 | + } |
| 30 | + |
| 31 | + private int calculateSize(TreeNode root, Map<TreeNode, Integer> sizes) { |
| 32 | + if (root == null) return 0; |
| 33 | + |
| 34 | + int left = calculateSize(root.left, sizes); |
| 35 | + int right = calculateSize(root.right, sizes); |
| 36 | + |
| 37 | + sizes.put(root, left + right + 1); |
| 38 | + return left + right + 1; |
| 39 | + } |
| 40 | + |
| 41 | + private int findKth(TreeNode root, int k, Map<TreeNode, Integer> sizes) { |
| 42 | + int left = (root.left == null) ? 0 : sizes.get(root.left); |
| 43 | + int right = (root.right == null) ? 0 : sizes.get(root.right); |
| 44 | + |
| 45 | + if (left == k - 1) { |
| 46 | + return root.val; |
| 47 | + } else if (left >= k) { |
| 48 | + return findKth(root.left, k, sizes); |
| 49 | + } else { |
| 50 | + return findKth(root.right, k - left - 1, sizes); |
| 51 | + } |
| 52 | + } |
| 53 | +} |
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