【leetcode】998. Maximum Binary Tree II
题目如下:
We are given the
root
node of a maximum tree: a tree where every node has a value greater than any other value in its subtree.Just as in the previous problem, the given tree was constructed from an list
A
(root = Construct(A)
) recursively with the followingConstruct(A)
routine:
- If
A
is empty, returnnull
.- Otherwise, let
A[i]
be the largest element ofA
. Create aroot
node with valueA[i]
.- The left child of
root
will beConstruct([A[0], A[1], ..., A[i-1]])
- The right child of
root
will beConstruct([A[i+1], A[i+2], ..., A[A.length - 1]])
- Return
root
.Note that we were not given A directly, only a root node
root = Construct(A)
.Suppose
B
is a copy ofA
with the valueval
appended to it. It is guaranteed thatB
has unique values.Return
Construct(B)
.
Example 1:
Input: root = [4,1,3,null,null,2], val = 5 Output: [5,4,null,1,3,null,null,2] Explanation: A = [1,4,2,3], B = [1,4,2,3,5]
Example 2:
Input: root = [5,2,4,null,1], val = 3 Output: [5,2,4,null,1,null,3] Explanation: A = [2,1,5,4], B = [2,1,5,4,3]
Example 3:
Input: root = [5,2,3,null,1], val = 4 Output: [5,2,4,null,1,3] Explanation: A = [2,1,5,3], B = [2,1,5,3,4]
Note:
1 <= B.length <= 100
解题思路:本题的题意需要好好理解一番,大概意思是这样。给定树A的根节点,并且A是一个最大二叉树(最大二叉树的定义见题目中previous problem),根据特性将A还原成一个数组的表示,如用例1是[1,4,2,3],在这个数组后面追加一个元素后变成[1,4,2,3,5],再根据这个新得的数组构造出树B。那么解题方法也可以分成两部,先是还原成数组,这个用递归即可;二是构造树,见【leetcode】654. Maximum Binary Tree 。
代码如下:
# 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 build(self, node, nums): v = max(nums) inx = nums.index(v) node.val = v ll = nums[:inx] rl = nums[inx + 1:] if len(ll) > 0: left = TreeNode(None) node.left = left self.build(left, ll) if len(rl) > 0: right = TreeNode(None) node.right = right self.build(right, rl) def decompile(self,node): if node == None: return [] return self.decompile(node.left) + [node.val] + self.decompile(node.right) def insertIntoMaxTree(self, root, val): """ :type root: TreeNode :type val: int :rtype: TreeNode """ path = self.decompile(root) + [val] newRoot = TreeNode(None) self.build(newRoot, path) return newRoot