construct binary tree from preorder and inorder traversal

这题当时也是没做出来,还是心不够静,太急于求成了。

解法1递归

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# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class (object):
def buildTree(self, preorder, inorder):
"""
:type preorder: List[int]
:type inorder: List[int]
:rtype: TreeNode
"""
if not inorder:
return None
root = TreeNode(preorder.pop(0))
index = inorder.index(root.val)
root.left = self.buildTree(preorder, inorder[:index])
root.right = self.buildTree(preorder, inorder[index+1:])
return root

解法2递归,不改变原有的array

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# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class (object):
def buildTree(self, preorder, inorder):
"""
:type preorder: List[int]
:type inorder: List[int]
:rtype: TreeNode
"""
return self.helper(preorder, inorder, 0, len(inorder) - 1, 0, len(inorder) - 1)
def helper(self, preorder, inorder, k, l, i, j):
if k > len(preorder) - 1 or i > j:
return None
rootValue = preorder[k]
root = TreeNode(rootValue)
inorderIndex = inorder.index(rootValue)
root.left = self.helper(preorder, inorder, k+1, k+inorderIndex-i-1, i, inorderIndex-1)
root.right = self.helper(preorder, inorder, k+inorderIndex+1-i, l, inorderIndex+1, j)
return root

最好再试一试非递归的解法