本文翻譯自：Difference between binary tree and binary search tree
任何人都可以用一個例子解釋二叉樹和二叉搜索樹之間的區別嗎？
#1樓
參考：https://stackoom.com/question/Qlmx/二叉樹和二叉搜索樹之間的區別
#2樓
A binary tree is made of nodes, where each node contains a "left" pointer, a "right" pointer, and a data element. 二叉樹由節點組成，其中每個節點包含“左”指針，“右”指針和數據元素。 The "root" pointer points to the topmost node in the tree. “根”指針指向樹中最頂層的節點。 The left and right pointers recursively point to smaller "subtrees" on either side. 左右指針遞歸地指向任一側較小的“子樹”。 A null pointer represents a binary tree with no elements -- the empty tree. 空指針表示沒有元素的二叉樹 - 空樹。 The formal recursive definition is: a binary tree is either empty (represented by a null pointer), or is made of a single node, where the left and right pointers (recursive definition ahead) each point to a binary tree. 形式遞歸定義是：二叉樹是空的（由空指針表示），或者由單個節點組成，其中左右指針（前面的遞歸定義）均指向二叉樹。
A binary search tree (BST) or "ordered binary tree" is a type of binary tree where the nodes are arranged in order: for each node, all elements in its left subtree are less to the node (<), and all the elements in its right subtree are greater than the node (>). 二叉搜索樹 （BST）或“有序二叉樹”是一種二叉樹，其中節點按順序排列：對於每個節點，其左子樹中的所有元素都少於節點（<），以及所有元素在其右子樹中大於節點（>）。
5
/ \
3 6
/ \ \
1 4 9
The tree shown above is a binary search tree -- the "root" node is a 5, and its left subtree nodes (1, 3, 4) are < 5, and its right subtree nodes (6, 9) are > 5. Recursively, each of the subtrees must also obey the binary search tree constraint: in the (1, 3, 4) subtree, the 3 is the root, the 1 < 3 and 4 > 3. 上面顯示的樹是二叉搜索樹 - “根”節點是5，其左子樹節點（1,3,4）是<5，其右子樹節點（6,9）是> 5。遞歸地，每個子樹還必須服從二元搜索樹約束：在（1,3,4）子樹中，3是根，1 <3和4> 3。
Watch out for the exact wording in the problems -- a "binary search tree" is different from a "binary tree". 注意問題中的確切措辭 - “二叉搜索樹”與“二叉樹”不同。
#3樓
二叉搜索樹是一種特殊的二叉樹，它具有以下屬性：對於任何節點n，n的左子樹中的每個後代節點的值小於n的值，並且右子樹中的每個後代節點的值是大於n的值。
#4樓
As everybody above has explained about the difference between binary tree and binary search tree, i am just adding how to test whether the given binary tree is binary search tree. 正如上面的每個人都解釋了二叉樹和二叉搜索樹之間的區別，我只是添加如何測試給定的二叉樹是否是二叉搜索樹。
boolean b = new Sample().isBinarySearchTree(n1, Integer.MIN_VALUE, Integer.MAX_VALUE);
.......
.......
.......
public boolean isBinarySearchTree(TreeNode node, int min, int max)
{
if(node == null)
{
return true;
}
boolean left = isBinarySearchTree(node.getLeft(), min, node.getValue());
boolean right = isBinarySearchTree(node.getRight(), node.getValue(), max);
return left && right && (node.getValue()<max) && (node.getValue()>=min);
}
Hope it will help you. 希望它會對你有所幫助。 Sorry if i am diverting from the topic as i felt it's worth mentioning this here. 對不起，如果我偏離主題，因爲我覺得這裏值得一提。
#5樓
Binary Tree is a specialized form of tree with two child (left child and right Child). 二叉樹是一種特殊形式的樹，有兩個孩子（左孩子和右孩子）。 It is simply representation of data in Tree structure 它只是Tree結構中數據的表示
Binary Search Tree (BST) is a special type of Binary Tree that follows following condition: 二進制搜索樹（BST）是一種特殊類型的二叉樹，遵循以下條件：
- left child node is smaller than its parent Node left子節點小於其父節點
- right child node is greater than its parent Node 右子節點大於其父節點
#6樓
A binary tree is a tree whose children are never more than two. 二叉樹是一棵樹，其子項永遠不會超過兩個。 A binary search tree follows the invariant that the left child should have a smaller value than the root node's key, while the right child should have a greater value than the root node's key. 二元搜索樹遵循不變量，即左子節點的值應小於根節點的鍵，而右子節點的值應大於根節點的鍵。