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Update 0106.从中序与后序遍历序列构造二叉树.md

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python 代码重复
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problems/0106.从中序与后序遍历序列构造二叉树.md
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@ -400,8 +400,6 @@ public:
}; };
``` ```
## Python
# 105.从前序与中序遍历序列构造二叉树 # 105.从前序与中序遍历序列构造二叉树
@ -657,38 +655,6 @@ class Solution {
## Python ## Python
```python
class Solution:
def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
# 第一步: 特殊情况讨论: 树为空. 或者说是递归终止条件
if not postorder:
return
# 第二步: 后序遍历的最后一个就是当前的中间节点
root_val = postorder[-1]
root = TreeNode(root_val)
# 第三步: 找切割点.
root_index = inorder.index(root_val)
# 第四步: 切割inorder数组. 得到inorder数组的左,右半边.
left_inorder = inorder[:root_index]
right_inorder = inorder[root_index + 1:]
# 第五步: 切割postorder数组. 得到postorder数组的左,右半边.
# ⭐️ 重点1: 中序数组大小一定跟后序数组大小是相同的.
left_postorder = postorder[:len(left_inorder)]
right_postorder = postorder[len(left_inorder): len(postorder) - 1]
# 第六步: 递归
root.left = self.buildTree(left_inorder, left_postorder)
root.right = self.buildTree(right_inorder, right_postorder)
# 第七步: 返回答案
return root
```
105.从前序与中序遍历序列构造二叉树 105.从前序与中序遍历序列构造二叉树
```python ```python
@ -717,7 +683,7 @@ class Solution:
# 第六步: 递归 # 第六步: 递归
root.left = self.buildTree(preorder_left, inorder_left) root.left = self.buildTree(preorder_left, inorder_left)
root.right = self.buildTree(preorder_right, inorder_right) root.right = self.buildTree(preorder_right, inorder_right)
# 第七步: 返回答案
return root return root
``` ```
@ -749,7 +715,7 @@ class Solution:
# 第六步: 递归 # 第六步: 递归
root.left = self.buildTree(inorder_left, postorder_left) root.left = self.buildTree(inorder_left, postorder_left)
root.right = self.buildTree(inorder_right, postorder_right) root.right = self.buildTree(inorder_right, postorder_right)
# 第七步: 返回答案
return root return root
``` ```