530

Author: ProHiryu

DFS/traditionalDFS/530.md

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+## 530 Minimum Absolute Difference in BST
+
+#### Description
+
+[link](https://leetcode.com/problems/minimum-absolute-difference-in-bst/)
+
+---
+
+#### Solution
+
+**这里讲一下DFS的时间复杂度判断：如果是最坏情况的话，首先因为初始DFS的循环是O(M + N)，每个点的最大遍历长度是O(MN)，这就是最坏情况的时间复杂度**
+
+---
+
+#### Code
+
+> 最坏情况下是O(n^2)
+
+Iteration:
+
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def getMinimumDifference(self, root: TreeNode) -> int:
+        stack = []
+        n = float('inf')
+        res = float('inf')
+        
+        while root or stack:
+            if root:
+                stack.append(root)
+                root = root.left
+            else:
+                root = stack.pop()
+                m = root.val
+                res = min(abs(m-n), res)
+                n = m
+                root = root.right
+        
+        return res
+```
+
+Recursive:
+
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def getMinimumDifference(self, root: TreeNode) -> int:
+        L = []
+        def dfs(node):
+            if node.left: dfs(node.left)
+            L.append(node.val)
+            if node.right: dfs(node.right)
+        dfs(root)
+        return min(b - a for a, b in zip(L, L[1:]))
+```