|
| 1 | +### 二叉树的前中后序遍历(递归) |
| 2 | +#### 前序遍历 |
| 3 | +``` |
| 4 | +vector <int> res; |
| 5 | + |
| 6 | +vector<int> main(TreeNode* root){ |
| 7 | + dfs(root); |
| 8 | + return res; |
| 9 | +} |
| 10 | + |
| 11 | +void dfs(TreeNode* root){ |
| 12 | + if (!root) return; |
| 13 | + res.push_back(root->val); |
| 14 | + dfs(root->left); |
| 15 | + dfs(root->right); |
| 16 | +} |
| 17 | +``` |
| 18 | +#### 中序遍历 |
| 19 | +``` |
| 20 | +vector <int> res; |
| 21 | + |
| 22 | +vector<int> main(TreeNode* root){ |
| 23 | + dfs(root); |
| 24 | + return res; |
| 25 | +} |
| 26 | + |
| 27 | +void dfs(TreeNode* root){ |
| 28 | + if (!root) return; |
| 29 | + dfs(root->left); |
| 30 | + res.push_back(root->val); |
| 31 | + dfs(root->right); |
| 32 | +} |
| 33 | +``` |
| 34 | +#### 后序遍历 |
| 35 | +``` |
| 36 | +vector <int> res; |
| 37 | + |
| 38 | +vector<int> main(TreeNode* root){ |
| 39 | + dfs(root); |
| 40 | + return res; |
| 41 | +} |
| 42 | + |
| 43 | +void dfs(TreeNode* root){ |
| 44 | + if (!root) return; |
| 45 | + dfs(root->left); |
| 46 | + dfs(root->right); |
| 47 | + res.push_back(root->val); |
| 48 | +} |
| 49 | +``` |
| 50 | +### 二叉树的前中后序遍历(迭代) |
| 51 | +#### 前序遍历 |
| 52 | +- 思路:二叉树的迭代遍历一般用栈来存储 |
| 53 | +栈是先进后出的顺序,如果是前序遍历,输出为中->左->右,所以先将右儿子入栈,再将左儿子入栈 |
| 54 | +- 模板解法: |
| 55 | + 1. 初始化栈 |
| 56 | + 2. 当前节点赋值为根节点 |
| 57 | + 3. 将根节点和所有左孩子入栈并加入答案直至当前节点为空 |
| 58 | + 4. 每弹出一个栈顶元素就到达当前节点的右孩子 |
| 59 | +- 代码模板 |
| 60 | +``` |
| 61 | +stack<TreeNode*> stk; |
| 62 | +vector<int> res; |
| 63 | +auto cur = root; |
| 64 | +while (cur || stk.size()){ |
| 65 | + while (cur){ |
| 66 | + res.push_back(cur->val); |
| 67 | + stk.push(cur); |
| 68 | + cur = cur->left; |
| 69 | + } |
| 70 | + auto tmp = stk.top(); |
| 71 | + stk.pop(); |
| 72 | + cur = tmp->right; |
| 73 | +} |
| 74 | +``` |
| 75 | +- 常规解法: |
| 76 | + 1. 初始化栈 |
| 77 | + 2. 将根节点入栈 |
| 78 | + 3. 当栈不为空将栈顶元素记录弹出加入答案 |
| 79 | + 4. 如果元素有右儿子则将右儿子入栈,如果有左儿子则将左儿子入栈 |
| 80 | +- 代码模板 |
| 81 | +``` |
| 82 | +stack<TreeNode*> stk; |
| 83 | +vector<int> res; |
| 84 | +if (root) stk.push(root); |
| 85 | +while (stk.size()){ |
| 86 | + auto c = stk.top(); |
| 87 | + stk.pop(); |
| 88 | + res.push_back(c->val); |
| 89 | + if (c->right) stk.push(c->right); |
| 90 | + if (c->left) stk.push(c->left); |
| 91 | +} |
| 92 | +``` |
| 93 | +#### 中序遍历 |
| 94 | +- 思路:与前序遍历的模板解法相似,区别是前序遍历中在找树的左底部的过程中记录答案,而中序遍历是找到左底部再开始记录答案 |
| 95 | +- 模板解法: |
| 96 | + 1. 初始化栈 |
| 97 | + 2. 当前节点赋值为根节点 |
| 98 | + 3. 将根节点和所有左孩子入栈直至当前节点为空 |
| 99 | + 4. 每弹出一个栈顶元素加入答案并到达当前节点的右孩子 |
| 100 | +- 代码模板 |
| 101 | +``` |
| 102 | +stack<TreeNode*> stk; |
| 103 | +vector<int> res; |
| 104 | +auto cur = root; |
| 105 | +while (cur || stk.size()){ |
| 106 | + while (cur){ |
| 107 | + stk.push(cur); |
| 108 | + cur = cur->left; |
| 109 | + } |
| 110 | + auto temp = stk.top(); |
| 111 | + stk.pop(); |
| 112 | + res.push_back(temp->val); |
| 113 | + cur = temp->right; |
| 114 | +} |
| 115 | +``` |
| 116 | +#### 后序遍历 |
| 117 | +- 思路:与前序遍历的模板解法类似 |
| 118 | +栈是先进后出的顺序,如果是后序遍历,输出为左->右->中,和前序遍历中->左->右相比,可以在前序遍历的基础上将输入改为中->右->左,再将答案数组反转即为左->右->中 |
| 119 | +- 模板解法: |
| 120 | + 1. 初始化栈 |
| 121 | + 2. 当前节点赋值为根节点 |
| 122 | + 3. 将根节点和所有右孩子入栈并加入答案直至当前节点为空 |
| 123 | + 4. 每弹出一个栈顶元素就到达当前节点的左孩子 |
| 124 | + 5. 反转答案数组 |
| 125 | +- 代码模板 |
| 126 | +``` |
| 127 | +stack<TreeNode*> stk; |
| 128 | +vector<int> res; |
| 129 | +auto cur = root; |
| 130 | +while (cur || stk.size()){ |
| 131 | + while (cur){ |
| 132 | + stk.push(cur); |
| 133 | + res.push_back(cur->val); |
| 134 | + cur = cur->right; |
| 135 | + } |
| 136 | + auto temp = stk.top(); |
| 137 | + stk.pop(); |
| 138 | + cur = temp->left; |
| 139 | +} |
| 140 | +reverse(res.begin(), res.end()); |
| 141 | +``` |
| 142 | +### 二叉树的层序遍历 |
| 143 | +- 思路:二叉树的前中后序遍历都是用深度优先搜索的方法(dfs)主要使用栈实现,层序遍历是广度优先搜索主要使用队列实现 |
| 144 | +- 常规解法: |
| 145 | + 1. 初始化队列 |
| 146 | + 2. 将根节点加入队列中 |
| 147 | + 3. 当队列不为空时,弹出队头元素,加入到答案中 |
| 148 | + 4. 如果左子树非空,左子树加入队列 |
| 149 | + 5. 如果右子树非空,右子树加入队列 |
| 150 | +- 代码模板 |
| 151 | +``` |
| 152 | +queue<TreeNode*> q; |
| 153 | +vector<int> res; |
| 154 | +q.push(root); |
| 155 | +while (q.size()){ |
| 156 | + int n = q.size(); |
| 157 | + for (int i = 0; i < n; i ++){ |
| 158 | + auto cur = q.front(); |
| 159 | + q.pop(); |
| 160 | + res.push_back(cur->val); |
| 161 | + if (cur->left) q.push(cur->left); |
| 162 | + if (cur->right) q.push(cur->right); |
| 163 | + } |
| 164 | +} |
| 165 | +``` |
| 166 | +### 二叉树有关的题目和二叉树遍历的关系 |
| 167 | +#### leetcode.226 |
| 168 | +- 链接<https://leetcode.cn/problems/invert-binary-tree/> |
| 169 | +- 解题方法:有三种方法解本道题 |
| 170 | + 从上到下遍历树的时候翻转左右节点->前序遍历 |
| 171 | + 遍历到树的叶子节点回溯的时候翻转左右节点->后序遍历 |
| 172 | + 遍历每一层,将每一层的左右节点翻转->层序遍历 |
| 173 | +- leetcode解题代码 |
| 174 | +``` |
| 175 | +class Solution { |
| 176 | +public: |
| 177 | + TreeNode* invertTree(TreeNode* root) { |
| 178 | + // 递归(终止条件) |
| 179 | + if (!root) return root; |
| 180 | + // 前序遍历(递归解法) |
| 181 | + swap(root->left, root->right); |
| 182 | + invertTree(root->left); |
| 183 | + invertTree(root->right); |
| 184 | + // 递归返回值 |
| 185 | + return root; |
| 186 | + } |
| 187 | +}; |
| 188 | + |
| 189 | +class Solution { |
| 190 | +public: |
| 191 | + TreeNode* invertTree(TreeNode* root) { |
| 192 | + // 递归(终止条件) |
| 193 | + if (!root) return root; |
| 194 | + // 后序遍历(递归解法) |
| 195 | + invertTree(root->left); |
| 196 | + invertTree(root->right); |
| 197 | + swap(root->left, root->right); |
| 198 | + // 递归返回值 |
| 199 | + return root; |
| 200 | + } |
| 201 | +}; |
| 202 | + |
| 203 | +class Solution { |
| 204 | +public: |
| 205 | + TreeNode* invertTree(TreeNode* root) { |
| 206 | + // 层序遍历模板 |
| 207 | + queue<TreeNode*> q; |
| 208 | + if (root) q.push(root); |
| 209 | + while (q.size()){ |
| 210 | + int n = q.size(); |
| 211 | + for (int i = 0; i < n; i ++){ |
| 212 | + auto c = q.front(); |
| 213 | + q.pop(); |
| 214 | + swap(c->left, c->right); |
| 215 | + if (c->left) q.push(c->left); |
| 216 | + if (c->right) q.push(c->right); |
| 217 | + } |
| 218 | + } |
| 219 | + return root; |
| 220 | + } |
| 221 | +}; |
| 222 | +``` |
| 223 | +#### leetcode.589 |
| 224 | +- 链接<https://leetcode.cn/problems/n-ary-tree-preorder-traversal/> |
| 225 | +- 解题方法:前序遍历递归模板或前序遍历迭代常规解法模板 |
| 226 | +- leetcode解题代码 |
| 227 | +``` |
| 228 | +class Solution { |
| 229 | +public: |
| 230 | + vector<int> res; |
| 231 | + vector<int> preorder(Node* root) { |
| 232 | + // 递归终止条件 |
| 233 | + if (!root) return res; |
| 234 | + // 前序遍历 |
| 235 | + res.push_back(root->val); |
| 236 | + for (auto c: root->children) preorder(c); |
| 237 | + return res; |
| 238 | + } |
| 239 | +}; |
| 240 | + |
| 241 | +class Solution { |
| 242 | +public: |
| 243 | + vector<int> preorder(Node* root) { |
| 244 | + // 与前序遍历常规解法相似 |
| 245 | + stack<Node*> stk; |
| 246 | + vector<int> res; |
| 247 | + if (root) stk.push(root); |
| 248 | + while (stk.size()){ |
| 249 | + auto c = stk.top(); |
| 250 | + stk.pop(); |
| 251 | + res.push_back(c->val); |
| 252 | + for (int i = c->children.size() - 1; i >= 0; i --){ |
| 253 | + stk.push(c->children[i]); |
| 254 | + } |
| 255 | + } |
| 256 | + return res; |
| 257 | + } |
| 258 | +}; |
| 259 | +``` |
| 260 | +#### leetcode.590 |
| 261 | +- 链接<https://leetcode.cn/problems/n-ary-tree-postorder-traversal/> |
| 262 | +- 解题方法:后序遍历递归模板或后序遍历迭代常规解法模板 |
| 263 | +- leetcode解题代码 |
| 264 | +``` |
| 265 | +class Solution { |
| 266 | +public: |
| 267 | + vector<int> res; |
| 268 | + vector<int> postorder(Node* root) { |
| 269 | + // 递归终止条件 |
| 270 | + if (!root) return res; |
| 271 | + // 后序遍历 |
| 272 | + for (auto c: root->children) postorder(c); |
| 273 | + res.push_back(root->val); |
| 274 | + return res; |
| 275 | + } |
| 276 | +}; |
| 277 | + |
| 278 | +class Solution { |
| 279 | +public: |
| 280 | + vector<int> postorder(Node* root) { |
| 281 | + // 与后序遍历常规解法类似 |
| 282 | + stack<Node*> stk; |
| 283 | + vector<int> res; |
| 284 | + if (root) stk.push(root); |
| 285 | + while (stk.size()){ |
| 286 | + auto c = stk.top(); |
| 287 | + stk.pop(); |
| 288 | + res.push_back(c->val); |
| 289 | + for (int i = 0; i < c->children.size(); i ++){ |
| 290 | + stk.push(c->children[i]); |
| 291 | + } |
| 292 | + } |
| 293 | + reverse(res.begin(), res.end()); |
| 294 | + return res; |
| 295 | + } |
| 296 | +}; |
| 297 | +``` |
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