Binary Tree Preorder Traversal

Author: int32bit

README.md

@@ -52,6 +52,7 @@
 + [136 Single Number(位运算)](algorithms/SingleNumber)
 + [141 Linked List Cycle(快慢指针法)](algorithms/LinkedListCycle)
 + [142 Linked List Cycle II](algorithms/LinkedListCycle2)
++ [144 Binary Tree Preorder Traversal(迭代法前序遍历](algorithms/BinaryTreePreorderTraversal)
 + [146 LRU Cache(空间换时间)](algorithms/LRUCache)
 + [147 Insertion Sort List(链表操作，插入排序)](algorithms/InsertionSortList)
 + [150 Evaluate Reverse Polish Notation(栈的应用)](algorithms/EvaluateReversePolishNotation)

algorithms/BinaryTreePreorderTraversal/README.md

@@ -0,0 +1,44 @@
+## Binary Tree Preorder Traversal
+
+Given a binary tree, return the preorder traversal of its nodes' values.
+
+For example:
+Given binary tree {1,#,2,3},
+```
+   1
+    \
+     2
+    /
+   3
+```
+
+return `[1,2,3]`.
+
+Note: Recursive solution is trivial, could you do it iteratively?
+
+## Solution
+
+前序遍历，使用递归，就太简单了。
+
+如何使用迭代的方式,其实就是使用栈模拟递归(注意递归(DFS)和栈的关系，队列和BFS的关系)
+
+先访问节点，然后压入右孩子，压入左孩子.
+```cpp
+vector<int> preorderTraversal(TreeNode *root) {
+	vector<int> result;
+	if (root == NULL)
+		return result;
+	stack<TreeNode *>s;
+	s.push(root);
+	while (!s.empty()) {
+		TreeNode *p = s.top();
+		s.pop();
+		result.push_back(p->val);
+		if (p->right)
+			s.push(p->right);
+		if (p->left)
+			s.push(p->left);
+	}
+	return result;
+}
+```

algorithms/BinaryTreePreorderTraversal/tree.cpp

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+#include <iostream>
+#include <vector>
+#include <string>
+#include <algorithm>
+#include <cstdio>
+#include <stack>
+using namespace std;
+struct TreeNode {
+	int val;
+	TreeNode *left;
+	TreeNode *right;
+	TreeNode(int x) : val(x), left(nullptr), right(nullptr){}
+};
+class Solution {
+public:
+	vector<int> preorderTraversal(TreeNode *root) {
+		vector<int> result;
+		if (root == NULL)
+			return result;
+		stack<TreeNode *>s;
+		s.push(root);
+		while (!s.empty()) {
+			TreeNode *p = s.top();
+			s.pop();
+			result.push_back(p->val);
+			if (p->right)
+				s.push(p->right);
+			if (p->left)
+				s.push(p->left);
+		}
+		return result;
+	}
+};
+TreeNode *mk_node(int val)
+{
+	return new TreeNode(val);
+}
+TreeNode *mk_child(TreeNode *root, TreeNode *left, TreeNode *right)
+{
+	root->left = left;
+	root->right = right;
+	return root;
+}
+TreeNode *mk_child(TreeNode *root, int left, int right)
+{
+	return mk_child(root, new TreeNode(left), new TreeNode(right));
+}
+TreeNode *mk_child(int root, int left, int right)
+{
+	return mk_child(new TreeNode(root), new TreeNode(left), new TreeNode(right));
+}
+int main(int argc, char **argv)
+{
+	Solution solution;
+	TreeNode *root = mk_child(10,5,15);
+	mk_child(root->right, 6, 20);
+	mk_child(root->left, 2, 3);
+	auto result = solution.preorderTraversal(root);
+	for_each(result.begin(), result.end(), [](int i){cout << i << " ";});
+	cout << endl;
+	return 0;
+}