python指標在遞回函式中沒有改變

這是我要解決的問題。但是，我無法弄清楚為什么 ans 的指標在遞回函式中發生了變化，但在 main 函式中仍然相同。誰能指出問題出在哪里？

https://leetcode.com/problems/find-a-corresponding-node-of-a-binary-tree-in-a-clone-of-that-tree/solution/

class Solution:
    def getTargetCopy(self, original: TreeNode, cloned: TreeNode, target: TreeNode) -> TreeNode:
        # the helper dfs function
        def dfs(root):
            if not root: return
            if root.val == target.val:
                ans = root
                return
            dfs(root.left)
            dfs(root.right)
        
        # main function
        ans = TreeNode(0)
        dfs(cloned)
        return ans

另外，我嘗試將 return ans 更改為 return ans.right 并且它有效。仍然不明白為什么它可以作業，但上面的不能作業

class Solution:
    def getTargetCopy(self, original: TreeNode, cloned: TreeNode, target: TreeNode) -> TreeNode:
        # the helper dfs function
        def dfs(root):
            if not root: return
            if root.val == target.val:
                ans.right = root
                return
            dfs(root.left)
            dfs(root.right)
        
        # main function
        ans = TreeNode(0)
        dfs(cloned)
        return ans.right

uj5u.com熱心網友回復：

因為ans在內部函式中是內部函式的區域。您將需要使用成員變數 ( self.ans) 或將其設為兩個函式中的全域變數。

uj5u.com熱心網友回復：

比較這三個例子：

In [32]: def outer():
    ...:     def inner():
    ...:         val = 3
    ...:         print(f"inner val: {val}")
    ...:     val = 0
    ...:     print(f"outer1 val: {val}")
    ...:     inner()
    ...:     print(f"outer2 val: {val}")
    ...:

In [33]: outer()
outer1 val: 0
inner val: 3
outer2 val: 0

In [34]: def outer():
    ...:     val = 0
    ...:     def inner():
    ...:         val = 3
    ...:         print(f"inner val: {val}")
    ...:
    ...:     print(f"outer1 val: {val}")
    ...:     inner()
    ...:     print(f"outer2 val: {val}")
    ...:

In [35]: outer()
outer1 val: 0
inner val: 3
outer2 val: 0

In [36]: def outer():
    ...:     def inner():
    ...:         nonlocal val
    ...:         val = 3
    ...:         print(f"inner val: {val}")
    ...:     val = 0
    ...:     print(f"outer1 val: {val}")
    ...:     inner()
    ...:     print(f"outer2 val: {val}")
    ...:

In [37]: outer()
outer1 val: 0
inner val: 3
outer2 val: 3

只有最后一個按您希望的方式作業。Python 變數范圍默認是本地的，您需要通過nonlocal宣告明確告訴它在封閉范圍內查找，否則您將定義一個內部函式本地的新變數。

見https://www.education.io/edpresso/the-legb-rule-in-python

所以這應該作業：

class Solution:
    def getTargetCopy(self, original: TreeNode, cloned: TreeNode, target: TreeNode) -> TreeNode:
        # the helper dfs function
        def dfs(root):
            nonlocal ans
            if not root: return
            if root.val == target.val:
                ans = root
                return
            dfs(root.left)
            dfs(root.right)
        
        # main function
        ans = TreeNode(0)
        dfs(cloned)
        return ans

評論中討論的另一個例子：

In [44]: def outer():
    ...:     def inner():
    ...:         val["a"] = 3
    ...:         print(f"inner val: {val}")
    ...:     val = {"a": 1}
    ...:     print(f"outer1 val: {val}")
    ...:     inner()
    ...:     print(f"outer2 val: {val}")
    ...:

In [45]: outer()
outer1 val: {'a': 1}
inner val: {'a': 3}
outer2 val: {'a': 3}

在這里我們可以看到nonlocal不需要宣告......對于屬性或專案訪問，Python 似乎會尋找封閉范圍來查找實體。

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