【144】二叉树的前序遍历

1、递归法：

class Solution {
public:void preorder(TreeNode* root, vector<int> &res){if(root == nullptr){return;}res.push_back(root->val);preorder(root->left, res);preorder(root->right, res);}vector<int> preorderTraversal(TreeNode* root) {vector<int> res;preorder(root, res);return res;}
};

2、迭代法

 

class Solution {
public:vector<int> preorderTraversal(TreeNode* root) {vector<int> res;if(root == nullptr){return res;}stack<TreeNode*> stk;TreeNode* node = root;while(!stk.empty() || node != nullptr){while(node != nullptr){stk.push(node);res.push_back(node->val);node = node->left;}node = stk.top();stk.pop();node = node->right;}return res;}
};

【94】二叉树的中序遍历

1、递归法

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

2、迭代法 

class Solution {
public:vector<int> inorderTraversal(TreeNode* root) {vector<int> res;if(root == nullptr){return res;}stack<TreeNode*> stk;TreeNode* node = root;while(!stk.empty() || node != nullptr){while(node != nullptr){stk.push(node);node = node->left;}node = stk.top();res.push_back(node->val);stk.pop();node = node->right;}return res;}
};

复杂度分析：

时间复杂度：O（N）每个结点会遍历一次且只遍历一次 

空间复杂度：O（N）栈至多会存放所有树节点

【145】二叉树的后序遍历

1、递归法

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

2、迭代法 

class Solution {
public:vector<int> postorderTraversal(TreeNode* root) {vector<int> res;if(root == nullptr){return res;}stack<TreeNode*> stk;TreeNode * prev = nullptr;while(!stk.empty() || root != nullptr){while(root != nullptr){stk.push(root);root = root->left;}root = stk.top();stk.pop();if(root->right == nullptr || root->right == prev){res.push_back(root->val);prev = root;root = nullptr;}else{stk.push(root);root = root->right;}}return res;}
};

思路：

从根节点开始遍历，并将根节点入栈，再遍历他的左子树，并依次入栈，直到该结点没有左子树。判断这个结点是否有右子树，如果没有，则将该结点弹出栈，并记录结点值。如果有则继续从他的右子树进行遍历，同时记录该结点的右子树是否遍历过，如果遍历过，则弹栈并记录结点值。