998. Maximum Binary Tree II-蒲公英云

998. Maximum Binary Tree II

A maximum tree is a tree where every node has a value greater than any other value in its subtree.

You are given the root of a maximum binary tree and an integer val.

Just as in the previous problem, the given tree was constructed from a 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 the 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]

题目：创建最大树，给定一颗最大二叉树，和一个值val, 将val以最右的形式插入二叉树。

思路：因为val是在构建二叉树的数组最后，因此在二叉树中往右子树找，找到最后一个比val大的节点。将其右子树改为新建节点的左子树，新建节点为其右子节点。代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* insertIntoMaxTree(TreeNode* root, int val) {
        TreeNode* node = new TreeNode(val);
        if(root && root->val < val) {
            node->left = root;
            return node;
        }
        TreeNode* nd = root;
        while(nd && nd->right && nd->right->val > val)
            nd = nd->right;
        node->left = nd->right;
        nd->right = node;
        return root;
    }
};