leetcode 98. Validate Binary Search Tree

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:
The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.

Example 1:

2
/ \
1   3

Binary tree 

[2,1,3]

,
return true.

Example 2:

1
/ \
2   3

Binary tree 

[1,2,3]

,
return false.

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/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isValidBST(TreeNode root) {
mTreeNode mroot=help(root);
help(mroot,root);
return mark;
}

boolean mark = true;

public mTreeNode help(TreeNode root) {
if(root==null)return null;
mTreeNode mroot=new mTreeNode(0,0);
if(root.left==null&&root.right==null){
mroot=new mTreeNode(root.val,root.val);
}
mroot.left=help(root.left);
mroot.right=help(root.right);
if(root.left!=null&&root.right==null){
mroot.min=Math.min(root.val,mroot.left.min);
mroot.max=Math.max(root.val,mroot.left.max);
}else if(root.left==null&&root.right!=null){
mroot.min=Math.min(root.val,mroot.right.min);
mroot.max=Math.max(root.val,mroot.right.max);
}else if(root.left!=null&&root.right!=null){
mroot.min=Math.min(Math.min(root.val,mroot.left.min),Math.min(root.val,mroot.right.min));

4000
mroot.max=Math.max(Math.max(root.val,mroot.left.max),Math.max(root.val,mroot.right.max));
}
return mroot;
}

public void help(mTreeNode mroot,TreeNode root){
if(mroot==null){return;}
if(mroot.left!=null){
if(mroot.left.max>=root.val){
mark=false;
return;
}
}
if(mroot.right!=null){
if(mroot.right.min<=root.val){
mark=false;
return;
}
}
help(mroot.left,root.left);
help(mroot.right,root.right);
}

static public class mTreeNode {
int min;
int max;
mTreeNode left;
mTreeNode right;

mTreeNode(int x,int y) {
min = x;
max=y;
}
}
}