Add solution 144

Author: yanglbme

README.md

@@ -51,6 +51,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 075 | [Sort Colors](https://github.com/yanglbme/leetcode/tree/master/solution/075.Sort%20Colors) | `Array`, `Two Pointers`, `Sort` |
 | 082 | [Remove Duplicates from Sorted List II](https://github.com/yanglbme/leetcode/tree/master/solution/082.Remove%20Duplicates%20from%20Sorted%20List%20II) | `Linked List` |
 | 094 | [Binary Tree Inorder Traversal](https://github.com/yanglbme/leetcode/tree/master/solution/094.Binary%20Tree%20Inorder%20Traversal) | `Hash Table`, `Stack`, `Tree` |
+| 144 | [Binary Tree Preorder Traversal](https://github.com/yanglbme/leetcode/tree/master/solution/144.Binary%20Tree%20Preorder%20Traversal) | `Stack`, `Tree` |
 | 153 | [Find Minimum in Rotated Sorted Array](https://github.com/yanglbme/leetcode/tree/master/solution/153.Find%20Minimum%20in%20Rotated%20Sorted%20Array) | `Array`, `Binary Search` |
 
 

solution/144.Binary Tree Preorder Traversal/README.md

@@ -0,0 +1,89 @@
+## 二叉树的前序遍历
+### 题目描述
+
+给定一个二叉树，返回它的 前序 遍历。
+
+ 示例:
+```
+输入: [1,null,2,3]  
+   1
+    \
+     2
+    /
+   3 
+
+输出: [1,2,3]
+```
+
+进阶: 递归算法很简单，你可以通过迭代算法完成吗？
+
+### 解法
+
+- 递归算法
+
+先访问根结点，再递归左子树，最后递归右子树。
+
+```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public List<Integer> preorderTraversal(TreeNode root) {
+        List<Integer> list = new ArrayList<>();
+        preorderTraversal(root, list);
+        return list;
+    }
+    
+    private void preorderTraversal(TreeNode root, List<Integer> list) {
+        if (root == null) {
+            return;
+        }
+        list.add(root.val);
+        preorderTraversal(root.left, list);
+        preorderTraversal(root.right, list);
+    }
+}
+```
+
+- 非递归算法
+循环：先访问根结点，接着判断是否有右孩子，有则压入栈中；再判断是否有左孩子，有则压入栈中。判断栈是否为空，若是，跳出循环。否则弹出栈顶元素，继续循环。
+
+```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public List<Integer> preorderTraversal(TreeNode root) {
+        List<Integer> list = new ArrayList<>();
+        Stack<TreeNode> stack = new Stack<>();
+        while (root != null) {
+            list.add(root.val);
+            if (root.right != null) {
+                stack.push(root.right);
+            }
+            if (root.left != null) {
+                stack.push(root.left);
+            }
+            if (!stack.isEmpty()) {
+                root = stack.pop();
+            } else {
+                break;
+            }
+        }
+        
+        return list;
+    }
+}
+```

solution/144.Binary Tree Preorder Traversal/Solution.java

@@ -0,0 +1,44 @@
+// 递归版本
+/*
+class Solution {
+    public List<Integer> preorderTraversal(TreeNode root) {
+        List<Integer> list = new ArrayList<>();
+        preorderTraversal(root, list);
+        return list;
+    }
+    
+    private void preorderTraversal(TreeNode root, List<Integer> list) {
+        if (root == null) {
+            return;
+        }
+        list.add(root.val);
+        preorderTraversal(root.left, list);
+        preorderTraversal(root.right, list);
+    }
+}
+
+*/
+
+// 非递归版本
+class Solution {
+    public List<Integer> preorderTraversal(TreeNode root) {
+        List<Integer> list = new ArrayList<>();
+        Stack<TreeNode> stack = new Stack<>();
+        while (root != null) {
+            list.add(root.val);
+            if (root.right != null) {
+                stack.push(root.right);
+            }
+            if (root.left != null) {
+                stack.push(root.left);
+            }
+            if (!stack.isEmpty()) {
+                root = stack.pop();
+            } else {
+                break;
+            }
+        }
+        
+        return list;
+    }
+}