Create 144. Binary Tree Preorder Traversal.md

solutions/144. Binary Tree Preorder Traversal.md

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+# [144. Binary Tree Preorder Traversal](https://leetcode.com/problems/binary-tree-preorder-traversal/)
+# 思路 
+要求进行二叉树的前序遍历, 属于务必掌握的基本题目.
+## 思路一 递归
+最简单的当然就是递归遍历了, 先遍历当前节点, 再递归遍历左子树, 再递归遍历右子树. 递归出口就是当前节点为空.
+
+## 思路二 迭代
+稍微难一点的就是迭代遍历. 这里一共有三种写法, 注意领会相互区别.
+
+# C++
+## 思路一 递归
+``` C++
+class Solution {
+private:
+    void helper(vector<int>&res, TreeNode *root){
+        if(!root) return;
+        res.push_back(root -> val);
+        helper(res, root -> left);
+        helper(res, root -> right);
+    }
+public:
+    vector<int> preorderTraversal(TreeNode* root) {
+        vector< int>res;
+        helper(res, root);
+        return res;
+    }
+};
+```
+
+## 思路二 迭代
+
+### 写法一
+``` C++
+class Solution {
+public:
+    vector<int> preorderTraversal(TreeNode* root) {
+        vector<int> res;
+        if (!root) return res;
+        stack<TreeNode*> stk{{root}}; // 注意这种初始化方法
+        TreeNode *p;
+        while (!stk.empty()) {
+            p = stk.top(); 
+            stk.pop();
+            
+            res.push_back(p -> val); // 先访问
+            if(p -> right) stk.push(p -> right); // 再将右子树入栈
+            if(p -> left) stk.push(p -> left);
+        }
+        return res;
+    }
+};
+```
+
+
+### 写法二
+``` C++
+class Solution {
+public:
+    vector<int> preorderTraversal(TreeNode* root) {
+        vector<int>res;
+        if(!root) return res;
+        
+        TreeNode *p = root;
+        stack<TreeNode *>stk;
+        while(!stk.empty() || p){
+            if(!p){ // 向左下降不动了, 从栈顶pop元素
+                p = stk.top();
+                stk.pop();
+            }
+            
+            res.push_back(p -> val);
+            if(p -> right) stk.push(p -> right); 
+            p = p -> left;
+        }
+        return res;
+    }
+};
+```
+
+
+### 写法三
+``` C++
+class Solution {
+public:
+    vector<int> preorderTraversal(TreeNode* root) {
+        vector<int>res;
+        if(!root) return res;
+        
+        TreeNode *p;
+        stack<TreeNode *>stk;
+        stk.push(root);
+        while(!stk.empty()){
+            p = stk.top();
+            stk.pop();
+            
+            while(p){ // 一路向左下降
+                res.push_back(p -> val);
+                if(p -> right) stk.push(p -> right); 
+                p = p -> left;
+            }
+        }
+        return res;
+    }
+};
+```