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> 利用二叉搜索树的特性搞起！

# 530.二叉搜索树的最小绝对差

[力扣题目链接](https://leetcode.cn/problems/minimum-absolute-difference-in-bst/)

给你一棵所有节点为非负值的二叉搜索树，请你计算树中任意两节点的差的绝对值的最小值。

示例：

![530二叉搜索树的最小绝对差](https://code-thinking-1253855093.file.myqcloud.com/pics/20201014223400123.png)

提示：树中至少有 2 个节点。

## 算法公开课

**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[二叉搜索树中，需要掌握如何双指针遍历！| LeetCode：530.二叉搜索树的最小绝对差](https://www.bilibili.com/video/BV1DD4y11779)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。


## 思路

题目中要求在二叉搜索树上任意两节点的差的绝对值的最小值。

**注意是二叉搜索树**，二叉搜索树可是有序的。

遇到在二叉搜索树上求什么最值啊，差值之类的，就把它想成在一个有序数组上求最值，求差值，这样就简单多了。

### 递归

那么二叉搜索树采用中序遍历，其实就是一个有序数组。

**在一个有序数组上求两个数最小差值，这是不是就是一道送分题了。**

最直观的想法，就是把二叉搜索树转换成有序数组，然后遍历一遍数组，就统计出来最小差值了。

代码如下：

```CPP
class Solution {
private:
vector<int> vec;
void traversal(TreeNode* root) {
    if (root == NULL) return;
    traversal(root->left);
    vec.push_back(root->val); // 将二叉搜索树转换为有序数组
    traversal(root->right);
}
public:
    int getMinimumDifference(TreeNode* root) {
        vec.clear();
        traversal(root);
        if (vec.size() < 2) return 0;
        int result = INT_MAX;
        for (int i = 1; i < vec.size(); i++) { // 统计有序数组的最小差值
            result = min(result, vec[i] - vec[i-1]);
        }
        return result;
    }
};
```

以上代码是把二叉搜索树转化为有序数组了，其实在二叉搜素树中序遍历的过程中，我们就可以直接计算了。

需要用一个pre节点记录一下cur节点的前一个节点。

如图：

![530.二叉搜索树的最小绝对差](https://code-thinking-1253855093.file.myqcloud.com/pics/20210204153247458.png)

一些同学不知道在递归中如何记录前一个节点的指针，其实实现起来是很简单的，大家只要看过一次，写过一次，就掌握了。

代码如下：

```CPP
class Solution {
private:
int result = INT_MAX;
TreeNode* pre = NULL;
void traversal(TreeNode* cur) {
    if (cur == NULL) return;
    traversal(cur->left);   // 左
    if (pre != NULL){       // 中
        result = min(result, cur->val - pre->val);
    }
    pre = cur; // 记录前一个
    traversal(cur->right);  // 右
}
public:
    int getMinimumDifference(TreeNode* root) {
        traversal(root);
        return result;
    }
};
```

是不是看上去也并不复杂！

### 迭代

看过这两篇[二叉树：听说递归能做的，栈也能做！](https://programmercarl.com/二叉树的迭代遍历.html)，[二叉树：前中后序迭代方式的写法就不能统一一下么？](https://programmercarl.com/二叉树的统一迭代法.html)文章之后，不难写出两种中序遍历的迭代法。

下面我给出其中的一种中序遍历的迭代法，代码如下：

```CPP
class Solution {
public:
    int getMinimumDifference(TreeNode* root) {
        stack<TreeNode*> st;
        TreeNode* cur = root;
        TreeNode* pre = NULL;
        int result = INT_MAX;
        while (cur != NULL || !st.empty()) {
            if (cur != NULL) { // 指针来访问节点，访问到最底层
                st.push(cur); // 将访问的节点放进栈
                cur = cur->left;                // 左
            } else {
                cur = st.top();
                st.pop();
                if (pre != NULL) {              // 中
                    result = min(result, cur->val - pre->val);
                }
                pre = cur;
                cur = cur->right;               // 右
            }
        }
        return result;
    }
};
```

## 总结

**遇到在二叉搜索树上求什么最值，求差值之类的，都要思考一下二叉搜索树可是有序的，要利用好这一特点。**

同时要学会在递归遍历的过程中如何记录前后两个指针，这也是一个小技巧，学会了还是很受用的。

后面我将继续介绍一系列利用二叉搜索树特性的题目。



## 其他语言版本


### Java 

递归
```java
class Solution {
    TreeNode pre;// 记录上一个遍历的结点
    int result = Integer.MAX_VALUE;
    public int getMinimumDifference(TreeNode root) {
       if(root==null)return 0;
       traversal(root);
       return result;
    }
    public void traversal(TreeNode root){
        if(root==null)return;
        //左
        traversal(root.left);
        //中
        if(pre!=null){
            result = Math.min(result,root.val-pre.val);
        }
        pre = root;
        //右
        traversal(root.right);
    }
}
```
統一迭代法-中序遍历
```Java
class Solution {
    public int getMinimumDifference(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode pre = null;
        int result = Integer.MAX_VALUE;

        if(root != null)
            stack.add(root);
        while(!stack.isEmpty()){
            TreeNode curr = stack.peek();
            if(curr != null){
                stack.pop();
                if(curr.right != null)
                    stack.add(curr.right);
                stack.add(curr);
                stack.add(null);
                if(curr.left != null)
                    stack.add(curr.left);
            }else{
                stack.pop();
                TreeNode temp = stack.pop();
                if(pre != null)
                    result = Math.min(result, temp.val - pre.val);
                pre = temp;
            }
        }
        return result;
    }
}
```

迭代法-中序遍历

```java
class Solution {
    TreeNode pre;
    Stack<TreeNode> stack;
    public int getMinimumDifference(TreeNode root) {
        if (root == null) return 0;
        stack = new Stack<>();
        TreeNode cur = root;
        int result = Integer.MAX_VALUE;
        while (cur != null || !stack.isEmpty()) {
            if (cur != null) {
                stack.push(cur); // 将访问的节点放进栈
                cur = cur.left; // 左
            }else {
                cur = stack.pop(); 
                if (pre != null) { // 中
                    result = Math.min(result, cur.val - pre.val);
                }
                pre = cur;
                cur = cur.right; // 右
            }
        }
        return result;
    }
}
```
### Python 

递归法（版本一）利用中序递增，结合数组
```python
class Solution:
    def __init__(self):
        self.vec = []

    def traversal(self, root):
        if root is None:
            return
        self.traversal(root.left)
        self.vec.append(root.val)  # 将二叉搜索树转换为有序数组
        self.traversal(root.right)

    def getMinimumDifference(self, root):
        self.vec = []
        self.traversal(root)
        if len(self.vec) < 2:
            return 0
        result = float('inf')
        for i in range(1, len(self.vec)):
            # 统计有序数组的最小差值
            result = min(result, self.vec[i] - self.vec[i - 1])
        return result

```



递归法（版本二）利用中序递增，找到该树最小值
```python 
class Solution:
    def __init__(self):
        self.result = float('inf')
        self.pre = None

    def traversal(self, cur):
        if cur is None:
            return
        self.traversal(cur.left)  # 左
        if self.pre is not None:  # 中
            self.result = min(self.result, cur.val - self.pre.val)
        self.pre = cur  # 记录前一个
        self.traversal(cur.right)  # 右

    def getMinimumDifference(self, root):
        self.traversal(root)
        return self.result

        
```

迭代法
```python 
class Solution:
    def getMinimumDifference(self, root):
        stack = []
        cur = root
        pre = None
        result = float('inf')

        while cur is not None or len(stack) > 0:
            if cur is not None:
                stack.append(cur)  # 将访问的节点放进栈
                cur = cur.left  # 左
            else:
                cur = stack.pop()
                if pre is not None:  # 中
                    result = min(result, cur.val - pre.val)
                pre = cur
                cur = cur.right  # 右

        return result


        
```
### Go

中序遍历，然后计算最小差值
```go
// 中序遍历的同时计算最小值
func getMinimumDifference(root *TreeNode) int {
    // 保留前一个节点的指针
    var prev *TreeNode
    // 定义一个比较大的值
    min := math.MaxInt64
    var travel func(node *TreeNode)
    travel = func(node *TreeNode) {
        if node == nil {
            return 
        }
        travel(node.Left)
        if prev != nil && node.Val - prev.Val < min {
            min = node.Val - prev.Val
        }
        prev = node
        travel(node.Right)
    }
    travel(root)
    return min
}
```

### JavaScript 
递归 先转换为有序数组
```javascript
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var getMinimumDifference = function (root) {
    let arr = [];
    const buildArr = (root) => {
        if (root) {
            buildArr(root.left);
            arr.push(root.val);
            buildArr(root.right);
        }
    }
    buildArr(root);
    let diff = arr[arr.length - 1];
    for (let i = 1; i < arr.length; ++i) {
        if (diff > arr[i] - arr[i - 1])
            diff = arr[i] - arr[i - 1];
    }
    return diff;
};
```
递归 在递归的过程中更新最小值
```js
var getMinimumDifference = function(root) {
    let res = Infinity
    let preNode = null
    // 中序遍历
    const inorder = (node) => {
        if(!node) return
        inorder(node.left)
        // 更新res
        if(preNode) res = Math.min(res, node.val - preNode.val)
        // 记录前一个节点         
        preNode = node
        inorder(node.right)
    }
    inorder(root)
    return res
}
```

迭代  中序遍历
```js
var getMinimumDifference = function(root) {
    let stack = []
    let cur = root
    let res = Infinity
    let pre = null
    while(cur || stack.length) {
        if(cur) {
            stack.push(cur)
            cur = cur.left
        } else {
            cur = stack.pop()
            if(pre) res = Math.min(res, cur.val - pre.val)
            pre = cur
            cur = cur.right
        }
    }
    return res
}
```

### TypeScript

> 辅助数组解决

```typescript
function getMinimumDifference(root: TreeNode | null): number {
    let helperArr: number[] = [];
    function recur(root: TreeNode | null): void {
        if (root === null) return;
        recur(root.left);
        helperArr.push(root.val);
        recur(root.right);
    }
    recur(root);
    let resMin: number = Infinity;
    for (let i = 0, length = helperArr.length; i < length - 1; i++) {
        resMin = Math.min(resMin, helperArr[i + 1] - helperArr[i]);
    }
    return resMin;
};
```

> 递归中解决

```typescript
function getMinimumDifference(root: TreeNode | null): number {
    let preNode: TreeNode | null= null;
    let resMin: number = Infinity;
    function recur(root: TreeNode | null): void {
        if (root === null) return;
        recur(root.left);
        if (preNode !== null) {
            resMin = Math.min(resMin, root.val - preNode.val);
        }
        preNode = root;
        recur(root.right);
    }
    recur(root);
    return resMin;
};
```

> 迭代法-中序遍历

```typescript
function getMinimumDifference(root: TreeNode | null): number {
    const helperStack: TreeNode[] = [];
    let curNode: TreeNode | null = root;
    let resMin: number = Infinity;
    let preNode: TreeNode | null = null;
    while (curNode !== null || helperStack.length > 0) {
        if (curNode !== null) {
            helperStack.push(curNode);
            curNode = curNode.left;
        } else {
            curNode = helperStack.pop()!;
            if (preNode !== null) {
                resMin = Math.min(resMin, curNode.val - preNode.val);
            }
            preNode = curNode;
            curNode = curNode.right;
        }
    }
    return resMin;
};
```

### Scala

构建二叉树的有序数组:

```scala
object Solution {
  import scala.collection.mutable
  def getMinimumDifference(root: TreeNode): Int = {
    val arr = mutable.ArrayBuffer[Int]()
    def traversal(node: TreeNode): Unit = {
      if (node == null) return
      traversal(node.left)
      arr.append(node.value)
      traversal(node.right)
    }
    traversal(root)
    // 在有序数组上求最小差值
    var result = Int.MaxValue
    for (i <- 1 until arr.size) {
      result = math.min(result, arr(i) - arr(i - 1))
    }
    result // 返回最小差值
  }
}
```

递归记录前一个节点:

```scala
object Solution {
  def getMinimumDifference(root: TreeNode): Int = {
    var result = Int.MaxValue // 初始化为最大值
    var pre: TreeNode = null // 记录前一个节点

    def traversal(cur: TreeNode): Unit = {
      if (cur == null) return
      traversal(cur.left)
      if (pre != null) {
        // 对比result与节点之间的差值
        result = math.min(result, cur.value - pre.value)
      }
      pre = cur
      traversal(cur.right)
    }

    traversal(root)
    result // return关键字可以省略
  }
}
```

迭代解决:

```scala
object Solution {
  import scala.collection.mutable
  def getMinimumDifference(root: TreeNode): Int = {
    var result = Int.MaxValue // 初始化为最大值
    var pre: TreeNode = null // 记录前一个节点
    var cur = root
    var stack = mutable.Stack[TreeNode]()
    while (cur != null || !stack.isEmpty) {
      if (cur != null) {
        stack.push(cur)
        cur = cur.left
      } else {
        cur = stack.pop()
        if (pre != null) {
          result = math.min(result, cur.value - pre.value)
        }
        pre = cur
        cur = cur.right
      }
    }
    result // return关键字可以省略
  }
}
```

### Rust

构建二叉树的有序数组:

```rust
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    pub fn get_minimum_difference(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut vec = vec![];
        Self::traversal(root, &mut vec);
        let mut min = i32::MAX;
        for i in 1..vec.len() {
            min = min.min(vec[i] - vec[i - 1])
        }
        min
    }
    pub fn traversal(root: Option<Rc<RefCell<TreeNode>>>, v: &mut Vec<i32>) {
        if root.is_none() {
            return;
        }
        let node = root.as_ref().unwrap().borrow();
        Self::traversal(node.left.clone(), v);
        v.push(node.val);
        Self::traversal(node.right.clone(), v);
    }
}
```

递归中解决

```rust
impl Solution {
    pub fn get_minimum_difference(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut pre = None;
        let mut min = i32::MAX;
        Self::inorder(root, &mut pre, &mut min);
        min
    }
    pub fn inorder(root: Option<Rc<RefCell<TreeNode>>>, pre: &mut Option<i32>, min: &mut i32) {
        if root.is_none() {
            return;
        }
        let node = root.as_ref().unwrap().borrow();
        Self::inorder(node.left.clone(), pre, min);
        if let Some(pre) = pre {
            *min = (node.val - *pre).min(*min);
        }
        *pre = Some(node.val);

        Self::inorder(node.right.clone(), pre, min);
    }
}
```

迭代

```rust
impl Solution {
    pub fn get_minimum_difference(mut root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        if root.is_none() {
            return 0;
        }
        let mut stack = vec![];
        let mut pre = -1;
        let mut res = i32::MAX;
        while root.is_some() || !stack.is_empty() {
            while let Some(node) = root {
                root = node.borrow().left.clone();
                stack.push(node);
            }

            let node = stack.pop().unwrap();

            if pre >= 0 {
                res = res.min(node.borrow().val - pre);
            }

            pre = node.borrow().val;

            root = node.borrow().right.clone();
        }
        res
    }
}
```
### C#
```csharp
// 递归
public class Solution
{
    public List<int> res = new List<int>();
    public int GetMinimumDifference(TreeNode root)
    {
        Traversal(root);
        return res.SelectMany((x, i) => res.Skip(i + 1).Select(y => Math.Abs(x - y))).Min();

    }
    public void Traversal(TreeNode root)
    {
        if (root == null) return;
        Traversal(root.left);
        res.Add(root.val);
        Traversal(root.right);
    }
}
```


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