998.Maximum Binary Tree II（最大二叉树 II）

Description

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).

最大树定义：一个树，其中每个节点的值都大于其子树中的任何其他值。

给出最大树的根节点 root。

就像之前的问题那样，给定的树是从表 A（root = Construct(A)）递归地使用下述 Construct(A) 例程构造的：

如果 A 为空，返回 null
否则，令 A[i] 作为 A 的最大元素。创建一个值为 A[i] 的根节点 root
root 的左子树将被构建为 Construct([A[0], A[1], ..., A[i-1]])
root 的右子树将被构建为 Construct([A[i+1], A[i+2], ..., A[A.length - 1]])
返回 root

请注意，我们没有直接给定 A，只有一个根节点 root = Construct(A).

假设 B 是 A 的副本，并附加值 val。保证 B 中的值是不同的。

返回 Construct(B)。

题目链接：https://leetcode.com/problems/maximum-binary-tree-ii/

Difficulty: medium

Example 1:

maximum-binary-tree-ii-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:

maximum-binary-tree-ii-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:

maximum-binary-tree-ii-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

分析

由题意知，对于任意子树，根节点大于所有子节点，且每次插入的节点中序遍历序列的尾部，则节点x应插入到某节点a的右子树上（a的值大于x的值），当前位置子树移到a的左子树上；
若x大于root的值，则x左根节点，root左x的左子树；
找到ro1、ro2，其中ro1的值大于x的值，ro2的值小于x的值，ro2是ro1的右子树，x做ro1的右节点，ro2做x的左节点；
若ro2为None，x做ro1的右节点；
返回root。

参考代码

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 insertIntoMaxTree(self, root, val):
    p=TreeNode(val)
    if(root==None):
        return p
    if(val>=root.val):
        p.left=root
        return p
    ro1=root
    ro2=ro1.right
    while(ro2!=None):
        if(val>ro2.val):
            ro1.right=p
            p.left=ro2
            return root
        else:
            ro1=ro2
            ro2=ro1.right

    ro1.right=p
    return root