知識點:二叉樹,遞歸
思路:
pre爲先序遍歷序列,tin爲中序遍歷序列
- pre的第一個元素爲根節點的值,tin以root爲分界點
- root的索引爲i
- tin前半部分[:i]爲左子樹,後半部分[i+1:]爲右子樹
- pre[1:i+1]爲左子樹,[i+1:]爲右子樹
- 遞歸構造
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
# 返回構造的TreeNode根節點
def reConstructBinaryTree(self, pre, tin):
# write code here
if not pre and not tin:
return None
if set(pre) != set(tin):
return None
root = TreeNode(pre[0])
i = tin.index(pre[0])
root.left = self.reConstructBinaryTree(pre[1:i+1], tin[:i]) # 左子樹
root.right = self.reConstructBinaryTree(pre[i+1:], tin[i+1:]) # 右子樹
return root