889.construct binary tree from preorder and postorder traversal

Return any binary tree that matches the given preorder and postorder traversals.

Values in the traversals pre and post are distinct positive integers.

Input: pre = [1,2,4,5,3,6,7], post = [4,5,2,6,7,3,1]

Output: [1,2,3,4,5,6,7]

• 1 <= pre.length == post.length <= 30
• pre[] and post[] are both permutations of 1, 2, …, pre.length.
• It is guaranteed an answer exists. If there exists multiple answers, you can return any of them.

1. 根据先序遍历pre和后续遍历post构建二叉树，常规递归方法；
2. 递归的结束条件是：pre长度为0返回None，长度为1，返回TreeNode(pre[0])
3. pre[0]作为当前过程的根节点，pre[1:indexof(pre[1])+2]作为左子树先序，post[:indexof(pre[1])+1]作为左子树后序，pre[indexof(pre[1])+2:]作为右子树先序，post[indexof(pre[1])+1：-1]作为右子树后序；
4. 如例：{[2,4,5],[4,5,2]}为左子树，{[3，6，7],[6,7,3]}为右子树，indexof(pre[1])为pre[1] (2)在post中的位置，为：2
5. 以此类推

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

class Solution:
def constructFromPrePost(self, pre, post):
    if(len(pre)==0):
        return None
    if(len(pre)==1):
        return TreeNode(pre[0])
    head=TreeNode(pre[0])
    index=post.index(pre[1])
    left=self.constructFromPrePost(pre[1:2+index], post[:index+1])
    right=self.constructFromPrePost(pre[2+index:], post[index+1:-1])
    head.left=left
    head.right=right
    return head

306 / 306 test cases passed.
Runtime: 56 ms