106. 从中序与后序遍历序列构造二叉树
Difficulty: 中等
根据一棵树的中序遍历与后序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
例如，给出
中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]
返回如下的二叉树：
3
/
9 20
/
15 7
Solution
Language: 全部题目
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
def buildHelper(pos_left, pos_right, in_left, in_right):
if pos_right < pos_left: return None
pos_root = pos_right
in_root = index[postorder[pos_root]]
root = TreeNode(postorder[pos_root])
left_subtree_size = in_root - in_left # 属于左子树的节点的数量
root.left = buildHelper(pos_left, pos_left+left_subtree_size-1, in_left, in_root-1)
root.right = buildHelper(pos_left+left_subtree_size, pos_right-1, in_root+1, in_right)
return root
n = len(inorder)
index = {e: i for i, e in enumerate(inorder)}
return buildHelper(0, n-1, 0, n-1)