前序遍历 preorder = [3,9,20,15,7] 中序遍历 inorder = [9,3,15,20,7]
返回如下:
3 / \ 9 20 / \ 15 7
前序第一个必为根节点,而在中序遍历中位于此节点左边的是它的左子树,右边的是右子树。而在前序遍历中,左右子树连续的排列在根节点后。于是可用递归解此题。
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
def build(preleft,preright,inoleft,inoright):
if preleft >preright:
return None
preroot = preleft
inoroot = inorder.index(preorder[preroot])
size = inoroot - inoleft
root = TreeNode(preorder[preroot])
root.left = build(preleft+1,preleft+size,inoleft,inoroot-1)
root.right = build(preleft+size+1,preright,inoroot+1,inoright)
return root
n = len(preorder)
return build(0,n-1,0,n-1)