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10 | 10 |
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11 | 11 | public class RecoverBinarySearchTree { |
12 | 12 |
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13 | | - private TreeNode first, second, prev; |
| 13 | + TreeNode first, second, prev; |
14 | 14 |
|
15 | | - /** |
16 | | - * TestCase |
17 | | - * [0, 1]是最重要且易错的 |
18 | | - */ |
19 | 15 | public void recoverTree(TreeNode root) { |
20 | | - if (root == null) { |
21 | | - return; |
22 | | - } |
23 | | - inorderTraversal(root); |
24 | | - int t = first.val; |
| 16 | + inorder(root); |
| 17 | + int val = first.val; |
25 | 18 | first.val = second.val; |
26 | | - second.val = t; |
| 19 | + second.val = val; |
27 | 20 | } |
28 | 21 |
|
29 | | - // 首先first是可以一次性确定的,second不能,可能要被多次覆盖,所以 |
30 | | - // 下面设置了second后没有立即终止,而是继续遍历,直到最后的那个才是真正的 |
31 | | - private void inorderTraversal(TreeNode node) { |
32 | | - if (node == null) { |
33 | | - return; |
34 | | - } |
35 | | - |
36 | | - inorderTraversal(node.left); |
37 | | - |
38 | | - if (prev != null) { |
39 | | - // 两个节点交换,肯定是大的放前面了,小的放后面了,所以遇到的第一个prev大于node的,first肯定是prev. |
40 | | - // 遇到的第一个node小于prev的,second肯定是node |
41 | | - if (first == null && prev.val > node.val) { |
42 | | - first = prev; |
43 | | - } |
44 | | - |
45 | | - // 这里可否换成second == null,不能,因为second要多次覆盖 |
46 | | - if (first != null && prev.val > node.val) { |
47 | | - second = node; |
48 | | - // 这里可否加上break |
| 22 | + private void inorder(TreeNode root) { |
| 23 | + while (root != null) { |
| 24 | + if (root.left == null) { |
| 25 | + detect(root); |
| 26 | + root = root.right; |
| 27 | + } else { |
| 28 | + TreeNode pre = root.left; |
| 29 | + for ( ; pre.right != null && pre.right != root; pre = pre.right); |
| 30 | + if (pre.right == null) { |
| 31 | + pre.right = root; |
| 32 | + root = root.left; |
| 33 | + } else { |
| 34 | + pre.right = null; |
| 35 | + detect(root); |
| 36 | + root = root.right; |
| 37 | + } |
49 | 38 | } |
50 | | - |
51 | | - // 以上两个if为何可能同时满足,参考[0,1] |
52 | 39 | } |
53 | | - |
54 | | - prev = node; |
55 | | - |
56 | | - inorderTraversal(node.right); |
57 | 40 | } |
58 | 41 |
|
59 | | - |
60 | | - // 可换成非递归的写法 |
61 | | - private void inorderTraverse(TreeNode root) { |
| 42 | + private void inorder2(TreeNode root) { |
62 | 43 | Stack<TreeNode> stack = new Stack<>(); |
63 | | - |
64 | 44 | while (!stack.isEmpty() || root != null) { |
65 | 45 | if (root != null) { |
66 | 46 | stack.push(root); |
67 | 47 | root = root.left; |
68 | 48 | } else { |
69 | 49 | root = stack.pop(); |
70 | | - |
71 | | - if (prev != null) { |
72 | | - if (first == null && root.val < prev.val) { |
73 | | - first = prev; |
74 | | - } |
75 | | - |
76 | | - if (first != null && root.val < prev.val) { |
77 | | - second = root; |
78 | | - } |
79 | | - } |
80 | | - |
81 | | - prev = root; |
82 | | - |
83 | | - root = root.right; |
84 | | - } |
85 | | - } |
86 | | - } |
87 | | - |
88 | | - private void inorderTraverse2(TreeNode root) { |
89 | | - TreeNode temp; |
90 | | - |
91 | | - while (root != null) { |
92 | | - if (root.left == null) { |
93 | | - helper(root); |
| 50 | + detect(root); |
94 | 51 | root = root.right; |
95 | | - } else { |
96 | | - temp = root.left; |
97 | | - while (temp.right != null && temp.right != root) { |
98 | | - temp = temp.right; |
99 | | - } |
100 | | - if (temp.right == null) { |
101 | | - temp.right = root; |
102 | | - root = root.left; |
103 | | - } else { |
104 | | - temp.right = null; |
105 | | - helper(root); |
106 | | - root = root.right; |
107 | | - } |
108 | 52 | } |
109 | 53 | } |
110 | 54 | } |
111 | 55 |
|
112 | | - private void helper(TreeNode node) { |
| 56 | + private void detect(TreeNode node) { |
113 | 57 | if (prev != null) { |
114 | 58 | if (first == null && prev.val > node.val) { |
115 | 59 | first = prev; |
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