|
64 | 64 |
|
65 | 65 | <!-- 这里可写通用的实现逻辑 -->
|
66 | 66 |
|
| 67 | +先构建图。 |
| 68 | + |
| 69 | +然后对于树中的每一个节点 u,我们通过深度优先搜索的方法,递归地搜索它的所有子节点 v,计算出以 v 为根的子树的节点数目和权值之和。在这之后,我们将子节点的值分别进行累加,就可以得到以 u 为根的子树的节点数目和权值之和。如果权值之和为零,那么以 u 为根的子树需要被删除,我们将其节点数目也置为零,作为结果返回到上一层。最终根节点 0 对应的节点数目即为答案。 |
| 70 | + |
67 | 71 | <!-- tabs:start -->
|
68 | 72 |
|
69 | 73 | ### **Python3**
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70 | 74 |
|
71 | 75 | <!-- 这里可写当前语言的特殊实现逻辑 -->
|
72 | 76 |
|
73 | 77 | ```python
|
74 |
| - |
| 78 | +class Solution: |
| 79 | + def deleteTreeNodes(self, nodes: int, parent: List[int], value: List[int]) -> int: |
| 80 | + def dfs(u): |
| 81 | + for v in g[u]: |
| 82 | + dfs(v) |
| 83 | + value[u] += value[v] |
| 84 | + counter[u] += counter[v] |
| 85 | + if value[u] == 0: |
| 86 | + counter[u] = 0 |
| 87 | + |
| 88 | + g = defaultdict(list) |
| 89 | + for i, p in enumerate(parent): |
| 90 | + if p != -1: |
| 91 | + g[p].append(i) |
| 92 | + counter = [1] * nodes |
| 93 | + dfs(0) |
| 94 | + return counter[0] |
75 | 95 | ```
|
76 | 96 |
|
77 | 97 | ### **Java**
|
78 | 98 |
|
79 | 99 | <!-- 这里可写当前语言的特殊实现逻辑 -->
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80 | 100 |
|
81 | 101 | ```java
|
| 102 | +class Solution { |
| 103 | + private Map<Integer, List<Integer>> g; |
| 104 | + private int[] counter; |
| 105 | + private int[] value; |
| 106 | + |
| 107 | + public int deleteTreeNodes(int nodes, int[] parent, int[] value) { |
| 108 | + g = new HashMap<>(); |
| 109 | + for (int i = 0; i < nodes; ++i) { |
| 110 | + if (parent[i] != -1) { |
| 111 | + g.computeIfAbsent(parent[i], k -> new ArrayList<>()).add(i); |
| 112 | + } |
| 113 | + } |
| 114 | + this.value = value; |
| 115 | + counter = new int[nodes]; |
| 116 | + Arrays.fill(counter, 1); |
| 117 | + dfs(0); |
| 118 | + return counter[0]; |
| 119 | + } |
| 120 | + |
| 121 | + private void dfs(int u) { |
| 122 | + for (int v : g.getOrDefault(u, Collections.emptyList())) { |
| 123 | + dfs(v); |
| 124 | + value[u] += value[v]; |
| 125 | + counter[u] += counter[v]; |
| 126 | + } |
| 127 | + if (value[u] == 0) { |
| 128 | + counter[u] = 0; |
| 129 | + } |
| 130 | + } |
| 131 | +} |
| 132 | +``` |
| 133 | + |
| 134 | +### **C++** |
| 135 | + |
| 136 | +```cpp |
| 137 | +class Solution { |
| 138 | +public: |
| 139 | + unordered_map<int, vector<int>> g; |
| 140 | + vector<int> counter; |
| 141 | + vector<int> value; |
| 142 | + |
| 143 | + int deleteTreeNodes(int nodes, vector<int>& parent, vector<int>& value) { |
| 144 | + for (int i = 0; i < nodes; ++i) |
| 145 | + if (parent[i] != -1) |
| 146 | + g[parent[i]].push_back(i); |
| 147 | + counter.resize(nodes, 1); |
| 148 | + this->value = value; |
| 149 | + dfs(0); |
| 150 | + return counter[0]; |
| 151 | + } |
| 152 | + |
| 153 | + void dfs(int u) { |
| 154 | + for (int v : g[u]) |
| 155 | + { |
| 156 | + dfs(v); |
| 157 | + value[u] += value[v]; |
| 158 | + counter[u] += counter[v]; |
| 159 | + } |
| 160 | + if (value[u] == 0) counter[u] = 0; |
| 161 | + } |
| 162 | +}; |
| 163 | +``` |
82 | 164 |
|
| 165 | +### **Go** |
| 166 | +
|
| 167 | +```go |
| 168 | +func deleteTreeNodes(nodes int, parent []int, value []int) int { |
| 169 | + g := make(map[int][]int) |
| 170 | + for i, p := range parent { |
| 171 | + if p != -1 { |
| 172 | + g[p] = append(g[p], i) |
| 173 | + } |
| 174 | + } |
| 175 | + counter := make([]int, nodes) |
| 176 | + for i := range counter { |
| 177 | + counter[i] = 1 |
| 178 | + } |
| 179 | + var dfs func(u int) |
| 180 | + dfs = func(u int) { |
| 181 | + if vs, ok := g[u]; ok { |
| 182 | + for _, v := range vs { |
| 183 | + dfs(v) |
| 184 | + value[u] += value[v] |
| 185 | + counter[u] += counter[v] |
| 186 | + } |
| 187 | + } |
| 188 | + if value[u] == 0 { |
| 189 | + counter[u] = 0 |
| 190 | + } |
| 191 | + } |
| 192 | + dfs(0) |
| 193 | + return counter[0] |
| 194 | +} |
83 | 195 | ```
|
84 | 196 |
|
85 | 197 | ### **...**
|
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