8 - Medium - 从前序与中序遍历序列构造二叉树

根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

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

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def buildTree(self, preorder, inorder):
        """
        :type preorder: List[int]
        :type inorder: List[int]
        :rtype: TreeNode
        """
        def build_tree(prel, prer, pre, inl, inr, ino, pre_ino_map):
            if prel > prer or inl > inr:
                return
            root = TreeNode(pre[prel])
            pre_index = pre_ino_map[pre[prel]]
            root.left = build_tree(prel + 1, prel + pre_index - inl, pre, inl, pre_index - 1, ino, pre_ino_map)
            root.right = build_tree(prel + pre_index - inl + 1, prer, pre, pre_index + 1, inr, ino, pre_ino_map)
            return root

        ino_map = {}
        for i in range(len(inorder)):
            ino_map[inorder[i]] = i
        root = build_tree(0, len(preorder) - 1, preorder, 0, len(inorder) - 1, inorder, ino_map)
        return root