二叉树的前序遍历

题目链接：二叉树的前序遍历

解题思路1：递归

代码如下：

    void preorder(vector<int>& res, TreeNode* root){if(root == nullptr) return;//遇到空节点就返回res.push_back(root->val);//先遍历根节点preorder(res, root->left);//再遍历左子树preorder(res, root->right);//最后遍历右子树}vector<int> preorderTraversal(TreeNode* root) {vector<int> res;preorder(res, root);return res;}

解题思路2：辅助栈

代码如下：

    vector<int> preorderTraversal(TreeNode* root) {vector<int> res;if(root == nullptr) return res;stack<TreeNode*> s;s.push(root);while(!s.empty()){TreeNode* cur = s.top();res.push_back(cur->val);s.pop();if(cur->right) s.push(cur->right);if(cur->left) s.push(cur->left);}return res;}

二叉树的中序遍历

题目链接：二叉树的中序遍历

解题思路1：递归

代码如下：

    void inorder(vector<int>& res, TreeNode* root){if(root == nullptr) return;inorder(res, root->left);res.push_back(root->val);inorder(res, root->right);}vector<int> inorderTraversal(TreeNode* root) {vector<int> res;inorder(res, root);return res;}

解题思路2：辅助栈

代码如下：

    vector<int> inorderTraversal(TreeNode* root) {vector<int> res;stack<TreeNode*> s;while(root!=nullptr || !s.empty()){//每次找到最左节点while(root != nullptr){s.push(root);root = root->left;}//访问该节点TreeNode* cur = s.top();res.push_back(cur->val);s.pop();//进入右节点root = cur->right;}return res;}

二叉树的后序遍历

题目链接：二叉树的后序遍历

解题思路1：递归

代码如下：

    void postorder(vector<int>& res, TreeNode* root) {if (root == nullptr) return;postorder(res, root->left);postorder(res, root->right);res.push_back(root->val);}vector<int> postorderTraversal(TreeNode* root) {vector<int> res;postorder(res, root);return res;}

解题思路2：辅助栈

代码如下：

    vector<int> postorderTraversal(TreeNode* root) {vector<int> res;stack<TreeNode*> s;TreeNode* pre = nullptr;while(root!=nullptr || !s.empty()){//找到最左边的节点while(root != nullptr){s.push(root);root = root->left;}TreeNode* cur = s.top();s.pop();if(cur->right==nullptr || cur->right == pre){res.push_back(cur->val);pre = cur;}else{s.push(cur);root = cur->right;}}return res;}