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Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
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下面的代码内存超了，看了人家的，改成了这样。
一定要搞清前序中序是什么意思。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = Noneclass Solution(object):def buildTree(self, preorder, inorder):if inorder:val = preorder.pop(0)pos = inorder.index(val)root = TreeNode(val)root.left = self.buildTree(preorder,inorder[:pos])root.right = self.buildTree(preorder,inorder[pos+1:])return root'''if not p:return None#print p,i,i.index(p[0])if len(p) == 1:return TreeNode(p[0])val = p[0]pos = i.index(val)root = TreeNode(val)root.left = self.buildTree(p[1:pos+1],i[0:pos])root.right = self.buildTree(p[pos+1:],i[pos+1:])return root'''""":type preorder: List[int]:type inorder: List[int]:rtype: TreeNode"""这篇关于105. Construct Binary Tree from Preorder and Ignorer Traversal【M】【35】【再来一遍】的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
