Input the results of preorder traversal and inorder traversal of a binary tree, please build the binary tree and return its root node.
Assume that neither the input pre-order traversal nor the in-order traversal results contain duplicate numbers.
Example 1:
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7] Output: [3,9,20,null,null,15,7]
Example 2:
Input: preorder = [-1], inorder = [-1] Output: [-1]
Idea: It is necessary to understand the positional relationship between the root node of the binary tree and the left and right nodes in the pre-order traversal and post-order traversal.
Pre-order traversal: root left and right; in-order traversal: left root right. This reverses the construction of the binary tree.
# 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, preorder: List[int], inorder: List[int]) -> TreeNode:
if len(preorder) == 0:
return None
root = TreeNode(preorder[0]) # 构建根节点
idx = inorder.index(root.val) # 左子树的节点数目（标记所在根节点的下标，在中序遍历中，第一个到根节点的前一个的长度为二叉树左子树的长度）
root.left = self.buildTree(preorder[1:idx+1] , inorder[0:idx])
# 递归左子树，其中左子树部分的前序遍历的切片为第二个节点到左子树长度+1（左闭右开区间）；中序遍历为，总的中序遍历切片的第一个节点到左子树长度处。
root.right = self.buildTree(preorder[idx+1:] ,inorder[idx+1:])
# 递归右子树，其中右子树的前序遍历的切片为从左子树长度+1处开始一直到尾；中序遍历为总的中序遍历的切片的左子树长度+1处开始一直到尾。
return root
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