998. Maximum Binary Tree II（最大二叉树II）

题目描述

方法思路

Q1:Why to the right and not to the left?
Always go right since new element will be inserted at the end of the list.
Q2:why if(root.val<v){
TreeNode node = new TreeNode(v);
node.left=root;
return node;
},rather than if(root.val<v){
TreeNode node = new TreeNode(v);
node.right=root;
return node;
};
Is it just follow the example?

Approach1: iterative
Search on the right, find the node that cur.val > val > cur.right.val
Then create new node TreeNode(val),
put old cur.right as node.left,
put node as new cur.right.

public TreeNode insertIntoMaxTree(TreeNode root, int val) {
        TreeNode node = new TreeNode(val), cur = root;
        if (root.val < val) {
            node.left = root;
            return node;
        }
        while (cur.right != null && cur.right.val > val) {
            cur = cur.right;
        }
        node.left = cur.right;
        cur.right = node;
        return root;
    }

Approach2:recursive
这道题目描述的不是很清楚感觉上。（tricky!）
The idea is to insert node to the right parent or right sub-tree of current node. Using recursion can achieve this:
If inserted value is greater than current node, the inserted goes to right parent
If inserted value is smaller than current node, we recursively re-cauculate right subtree

class Solution {
	//Runtime: 2 ms, faster than 100.00%
    //Memory Usage: 37 MB, less than 100.00% 
    public TreeNode insertIntoMaxTree(TreeNode root, int v) {
        if(root==null)return new TreeNode(v);
        if(root.val<v){
            TreeNode node = new TreeNode(v);
            node.left=root;
            return node;
        }
        root.right=insertIntoMaxTree(root.right,v);
        return root;
    }
}