【leetcode】二叉树-从前序与中序遍历序列构造二叉树

第一种方法

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def buildTree(self, preorder, inorder):
        """
        :type preorder: List[int]
        :type inorder: List[int]
        :rtype: TreeNode
        """

        def bt(pre, ino):
            if len(pre)==0 or len(ino)==0:
                return None
            else:
                root = TreeNode(val=pre[0])
                for i in range(0, len(ino)):
                    if ino[i]== pre[0] :
                        in_left = ino[:i]
                        in_right = ino[(i+1):]
                        pre_left = pre[1:(len(in_left)+1)]
                        pre_right = pre[(len(in_left)+1):]
                        root.left = bt(pre_left, in_left)
                        root.right = bt(pre_right, in_right)
                        break
                return root
        return bt(preorder, inorder)         

第二种方法更简洁

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def buildTree(self, preorder, inorder):
        """
        :type preorder: List[int]
        :type inorder: List[int]
        :rtype: TreeNode
        """
        if len(preorder)==0 or len(inorder)==0:
             return None
        root = TreeNode(val=preorder[0])
        idx_root = inorder.index(preorder[0])
        root.left = self.buildTree(preorder[1:idx_root+1], inorder[:idx_root])
        root.right = self.buildTree(preorder[idx_root+1:], inorder[idx_root+1:])
        return root

一个问题

超出内存限制的原因：
在循环里有打印变量的语句（方便自己调试），但LeetCode测试用例变得很大时，这个打印语句就输出了大量内容，导致超出输出限制，注释掉多余打印语句即解决。
参考：https://www.zhihu.com/question/354716473