[Leetcode]Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

根据中序遍历和后序遍历构造二叉树~第一回写的代码如下：

class Solution:    # @param inorder, a list of integers    # @param postorder, a list of integers    # @return a tree node    def buildTree(self, inorder, postorder):        if not inorder: return None # inorder is empty        root = TreeNode(postorder[-1])        rootPos = inorder.index(postorder[-1])        root.left = self.buildTree(inorder[ : rootPos], postorder[ : rootPos])        root.right = self.buildTree(inorder[rootPos + 1 : ], postorder[rootPos : -1])        return root

但是一直Memory Limit Exceeded~后参照别人的代码，改进成如下才AC~在找中序遍历的root index时改用了dictionary来查找（而不是直接用list.index()），这种改进把算法的时间复杂度从O(n ^ 2) 提升到O(n)~还有就是注意一下细节~

class Solution:    # @param inorder, a list of integers    # @param postorder, a list of integers    # @return a tree node    def buildTree(self, inorder, postorder):        if inorder is None or postorder is None or len(inorder) == 0 or len(postorder) == 0:            return None        dict = {}        for i in xrange(len(inorder)):            dict[inorder[i]] = i        return self.helper(inorder, postorder, dict, 0, len(inorder) - 1, 0, len(inorder) - 1)            def helper(self, inorder, postorder, dict, inL, inR, postL, postR):        if inL > inR or postL > postR: return None        root = TreeNode(postorder[postR])        index = dict[postorder[postR]]        root.left = self.helper(inorder, postorder, dict, inL, index - 1, postL, postL + index - inL - 1)        root.right = self.helper(inorder, postorder, dict, index + 1, inR, postR - inR + index ,postR - 1)        return root