106.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.

# 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, inorder, postorder):
        """
        :type inorder: List[int]
        :type postorder: List[int]
        :rtype: TreeNode
        """
        if inorder and postorder:
            postorder.reverse()
            self.index = 0
            d = {}
            for i in xrange(0, len(inorder)):
                d[inorder[i]] = i
            return self.dfs(inorder, postorder, 0, len(postorder) - 1, d)
            
    def dfs(self, inorder, postorder, start, end, d):
        if start <= end:
            root = TreeNode(postorder[self.index])
            mid = d[postorder[self.index]]
            self.index += 1
            root.right = self.dfs(inorder, postorder, mid + 1, end, d)
            root.left = self.dfs(inorder, postorder, start, mid - 1, d)
            return root