从中序与后序遍历序列构造二叉树

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

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

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

我们发现在后序遍历中pop出来的结果是整个树的节点 比如 3， 20 然后找出这个节点在先序排列数组里面的位置 那么此位置的左边就是此节点的左子树 右边就是她的右子树.
步骤：
1. 从后序遍历里面 pop一个元素
2. 在先序遍历里面找到这个元素的位置 并且把他当做一个节点
3. 那么这个节点的左边就是左子树 右边就是右子树

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

class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
        def helper(in_left, in_right):
            # if there is no elements to construct subtrees
            if in_left > in_right:
                return None
            
            # pick up the last element as a root
            val = postorder.pop()
            root = TreeNode(val)

            # root splits inorder list
            # into left and right subtrees
            index = idx_map[val]
 
            # build right subtree
            root.right = helper(index + 1, in_right)
            # build left subtree
            root.left = helper(in_left, index - 1)
            return root
        
        # build a hashmap value -> its index
        idx_map = {val:idx for idx, val in enumerate(inorder)} 
        return helper(0, len(inorder) - 1)

 

吐槽： 用c做真的很难啊 还得自己做哈希表。。。。高级语言真的香