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).
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]
Constraints:
- 1 <= B.length <= 100
分析
题目的意思是：给你一颗最大二叉树，其中根结点是最大值。这道题我用递归的方法试了一下，发现总是有case ac不了。这里借鉴了别人的解法，如果为空，则直接建立一个结点并返回就行了。如果当前节点的值小于给定的val，说明找到了要插入的地方，创建一个结点t，t的左结点指向当前的结点，然后返回就行了。其他的就向右递归遍历就行了，不要问我为啥要一直向右递归，其实我也想知道，哈哈哈。我目测是找规律发现插入的位置都在右边，所以…
代码
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def insertIntoMaxTree(self, root: TreeNode, val: int) -> TreeNode:
if(root is None):
return TreeNode(val)
if(val>root.val):
t=TreeNode(val)
t.left=root
return t
root.right=self.insertIntoMaxTree(root.right,val)
return root