|
58 | 58 | <li><code>fromNode.right</code> 在测试用例的树中建立后为 <code>null</code> 。</li> |
59 | 59 | </ul> |
60 | 60 |
|
61 | | - |
62 | 61 | ## 解法 |
63 | 62 |
|
64 | 63 | <!-- 这里可写通用的实现逻辑 --> |
|
69 | 68 |
|
70 | 69 | <!-- 这里可写当前语言的特殊实现逻辑 --> |
71 | 70 |
|
| 71 | +记录父节点。 |
| 72 | + |
72 | 73 | ```python |
| 74 | +# Definition for a binary tree node. |
| 75 | +# class TreeNode: |
| 76 | +# def __init__(self, val=0, left=None, right=None): |
| 77 | +# self.val = val |
| 78 | +# self.left = left |
| 79 | +# self.right = right |
| 80 | +class Solution: |
| 81 | + def correctBinaryTree(self, root: TreeNode) -> TreeNode: |
| 82 | + q = collections.deque([root]) |
| 83 | + res = root |
| 84 | + p = {} |
| 85 | + while q: |
| 86 | + n = len(q) |
| 87 | + mp = {} |
| 88 | + for _ in range(n): |
| 89 | + node = q.popleft() |
| 90 | + if node.val in mp: |
| 91 | + left, father = p[mp[node.val]] |
| 92 | + if left: |
| 93 | + father.left = None |
| 94 | + else: |
| 95 | + father.right = None |
| 96 | + return res |
| 97 | + if node.left: |
| 98 | + q.append(node.left) |
| 99 | + p[node.left.val] = [True, node] |
| 100 | + if node.right: |
| 101 | + q.append(node.right) |
| 102 | + p[node.right.val] = [False, node] |
| 103 | + mp[node.right.val] = node.val |
| 104 | + return res |
| 105 | +``` |
73 | 106 |
|
| 107 | +优化，无需记录父节点。 |
| 108 | + |
| 109 | +```python |
| 110 | +# Definition for a binary tree node. |
| 111 | +# class TreeNode: |
| 112 | +# def __init__(self, val=0, left=None, right=None): |
| 113 | +# self.val = val |
| 114 | +# self.left = left |
| 115 | +# self.right = right |
| 116 | +class Solution: |
| 117 | + def correctBinaryTree(self, root: TreeNode) -> TreeNode: |
| 118 | + q = collections.deque([root]) |
| 119 | + while q: |
| 120 | + n = len(q) |
| 121 | + for _ in range(n): |
| 122 | + node = q.popleft() |
| 123 | + if node.right: |
| 124 | + if node.right.right in q: |
| 125 | + node.right = None |
| 126 | + return root |
| 127 | + q.append(node.right) |
| 128 | + if node.left: |
| 129 | + if node.left.right in q: |
| 130 | + node.left = None |
| 131 | + return root |
| 132 | + q.append(node.left) |
| 133 | + return root |
74 | 134 | ``` |
75 | 135 |
|
76 | 136 | ### **Java** |
77 | 137 |
|
78 | 138 | <!-- 这里可写当前语言的特殊实现逻辑 --> |
79 | 139 |
|
80 | 140 | ```java |
| 141 | +/** |
| 142 | + * Definition for a binary tree node. |
| 143 | + * public class TreeNode { |
| 144 | + * int val; |
| 145 | + * TreeNode left; |
| 146 | + * TreeNode right; |
| 147 | + * TreeNode() {} |
| 148 | + * TreeNode(int val) { this.val = val; } |
| 149 | + * TreeNode(int val, TreeNode left, TreeNode right) { |
| 150 | + * this.val = val; |
| 151 | + * this.left = left; |
| 152 | + * this.right = right; |
| 153 | + * } |
| 154 | + * } |
| 155 | + */ |
| 156 | +class Solution { |
| 157 | + public TreeNode correctBinaryTree(TreeNode root) { |
| 158 | + Deque<TreeNode> q = new ArrayDeque<>(); |
| 159 | + q.offer(root); |
| 160 | + while (!q.isEmpty()) { |
| 161 | + int n = q.size(); |
| 162 | + while (n-- > 0) { |
| 163 | + TreeNode node = q.pollFirst(); |
| 164 | + if (node.right != null) { |
| 165 | + if (node.right.right != null && q.contains(node.right.right)) { |
| 166 | + node.right = null; |
| 167 | + return root; |
| 168 | + } |
| 169 | + q.offer(node.right); |
| 170 | + } |
| 171 | + if (node.left != null) { |
| 172 | + if (node.left.right != null && q.contains(node.left.right)) { |
| 173 | + node.left = null; |
| 174 | + return root; |
| 175 | + } |
| 176 | + q.offer(node.left); |
| 177 | + } |
| 178 | + } |
| 179 | + } |
| 180 | + return root; |
| 181 | + } |
| 182 | +} |
| 183 | +``` |
81 | 184 |
|
| 185 | +### **C++** |
| 186 | + |
| 187 | +```cpp |
| 188 | +/** |
| 189 | + * Definition for a binary tree node. |
| 190 | + * struct TreeNode { |
| 191 | + * int val; |
| 192 | + * TreeNode *left; |
| 193 | + * TreeNode *right; |
| 194 | + * TreeNode() : val(0), left(nullptr), right(nullptr) {} |
| 195 | + * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} |
| 196 | + * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} |
| 197 | + * }; |
| 198 | + */ |
| 199 | +class Solution { |
| 200 | +public: |
| 201 | + TreeNode* correctBinaryTree(TreeNode* root) { |
| 202 | + queue<TreeNode*> q; |
| 203 | + q.push(root); |
| 204 | + unordered_set<TreeNode*> s; |
| 205 | + while (!q.empty()) |
| 206 | + { |
| 207 | + int n = q.size(); |
| 208 | + while (n--) |
| 209 | + { |
| 210 | + TreeNode* node = q.front(); |
| 211 | + q.pop(); |
| 212 | + if (node->right) |
| 213 | + { |
| 214 | + if (s.count(node->right->right)) |
| 215 | + { |
| 216 | + node->right = nullptr; |
| 217 | + return root; |
| 218 | + } |
| 219 | + q.push(node->right); |
| 220 | + s.insert(node->right); |
| 221 | + } |
| 222 | + if (node->left) |
| 223 | + { |
| 224 | + if (s.count(node->left->right)) |
| 225 | + { |
| 226 | + node->left = nullptr; |
| 227 | + return root; |
| 228 | + } |
| 229 | + q.push(node->left); |
| 230 | + s.insert(node->left); |
| 231 | + } |
| 232 | + } |
| 233 | + } |
| 234 | + return root; |
| 235 | + } |
| 236 | +}; |
82 | 237 | ``` |
83 | 238 |
|
84 | 239 | ### **...** |
|
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