• 【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 following Construct(A) routine:

    • If A is empty, return null.
    • Otherwise, let A[i] be the largest element of A.  Create a root node with value A[i].
    • The left child of root will be Construct([A[0], A[1], ..., A[i-1]])
    • The right child of root will be Construct([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 of A with the value val appended to it.  It is guaranteed that B 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. 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
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  • 原文地址:https://www.cnblogs.com/seyjs/p/10428480.html
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