【Leetcode刷题Python】从列表list中创建一颗二叉树

1 思路

使用递归创建二叉树，每个节点最多有两个子节点，以索引去从列表中取子节点的值，i为根节点，左子节点的值为2×i+1，右子节点的值为2×i+2。

2 python实现

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
def createTree(root,rt,i):
    if i <len(rt):
        if not rt[i]:
            return None
        else:
            root = TreeNode(rt[i])
            root.left =  createTree(root.left,rt,2*i+1)
            root.right = createTree(root.right,rt,2*i+2)
            return root
    return root 

root_list = [3,9,20,None,None,15,7]
root = TreeNode(-1)
root = createTree(root,root_list,0)