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7 | 7 |
|
8 | 8 | /** |
9 | 9 | * 111. Minimum Depth of Binary Tree |
| 10 | + * |
10 | 11 | * Given a binary tree, find its minimum depth. |
11 | 12 | * The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. |
12 | 13 | * */ |
13 | 14 |
|
14 | 15 | public class _111 { |
15 | | - /** |
16 | | - * We can solve this problem using both BFS and DFS: |
17 | | - * DFS is to visit every single root to leaf path and return the shortest one. |
18 | | - * BFS is to visit every level and return whenever we find the first leaf node. |
19 | | - */ |
20 | | - |
21 | | - public static class DFSSolution { |
22 | | - |
23 | | - public int minDepth(TreeNode root) { |
24 | | - if (root == null) { |
25 | | - return 0; |
26 | | - } |
27 | | - int left = minDepth(root.left); |
28 | | - int right = minDepth(root.right); |
29 | | - if (left == 0) { |
30 | | - return right + 1; |
31 | | - } |
32 | | - if (right == 0) { |
33 | | - return left + 1; |
34 | | - } |
35 | | - return Math.min(left, right) + 1; |
| 16 | + public static class Solution1 { |
| 17 | + /**DFS*/ |
| 18 | + public int minDepth(TreeNode root) { |
| 19 | + if (root == null) { |
| 20 | + return 0; |
36 | 21 | } |
37 | | - |
| 22 | + int left = minDepth(root.left); |
| 23 | + int right = minDepth(root.right); |
| 24 | + if (left == 0) { |
| 25 | + return right + 1; |
| 26 | + } |
| 27 | + if (right == 0) { |
| 28 | + return left + 1; |
| 29 | + } |
| 30 | + return Math.min(left, right) + 1; |
| 31 | + } |
38 | 32 | } |
39 | 33 |
|
40 | | - public static class BFSSolution { |
41 | | - |
42 | | - public int minDepth_BFS(TreeNode root) { |
43 | | - if (root == null) { |
44 | | - return 0; |
45 | | - } |
46 | | - Queue<TreeNode> q = new LinkedList(); |
47 | | - q.offer(root); |
48 | | - int level = 0; |
49 | | - while (!q.isEmpty()) { |
50 | | - level++; |
51 | | - int size = q.size(); |
52 | | - for (int i = 0; i < size; i++) { |
53 | | - TreeNode curr = q.poll(); |
54 | | - if (curr.left != null) { |
55 | | - q.offer(curr.left); |
56 | | - } |
57 | | - if (curr.right != null) { |
58 | | - q.offer(curr.right); |
59 | | - } |
60 | | - if (curr.left == null && curr.right == null) { |
61 | | - return level; |
62 | | - } |
63 | | - } |
64 | | - } |
| 34 | + public static class Solution2 { |
| 35 | + /**BFS*/ |
| 36 | + public int minDepth(TreeNode root) { |
| 37 | + if (root == null) { |
| 38 | + return 0; |
| 39 | + } |
| 40 | + Queue<TreeNode> q = new LinkedList(); |
| 41 | + q.offer(root); |
| 42 | + int level = 0; |
| 43 | + while (!q.isEmpty()) { |
| 44 | + level++; |
| 45 | + int size = q.size(); |
| 46 | + for (int i = 0; i < size; i++) { |
| 47 | + TreeNode curr = q.poll(); |
| 48 | + if (curr.left != null) { |
| 49 | + q.offer(curr.left); |
| 50 | + } |
| 51 | + if (curr.right != null) { |
| 52 | + q.offer(curr.right); |
| 53 | + } |
| 54 | + if (curr.left == null && curr.right == null) { |
65 | 55 | return level; |
| 56 | + } |
66 | 57 | } |
| 58 | + } |
| 59 | + return level; |
67 | 60 | } |
| 61 | + } |
68 | 62 | } |
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