Problem

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).

Example1

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]

Example2

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]

Example3

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]

Solution

通过中序遍历树，可以恢复出A，进而得到B，然后以B为输入，构造结果。

还可以直接在树上进行操作。
根据新加入的值是大于根节点还是小于根节点，分两种情况进行处理。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* insertIntoMaxTree(TreeNode* root, int val) {
        if(!root)
            return new TreeNode(val);
        
        TreeNode * prev = NULL;
        insertion(root,val,prev);
        if(prev)
            return prev;
        else
            return root;

    }

    void insertion(TreeNode* root, int val,TreeNode *&prev)
    {
        if(!root)
        {
            TreeNode * node = new TreeNode(val);
            if(prev)
                prev->right = node;
            return;
        }
        int rootVal = root->val;
        if(val > rootVal)
        {
            TreeNode *newRoot = new TreeNode(val);
            newRoot->left = root;
            if(prev)
                prev->right = newRoot;
            else
                prev = newRoot;
        }
        else
        {
            insertion(root->right,val,root);
        }
    }
};