largestBSTSubtree

#include "stdafx.h"#include <iostream>#include "limits.h"using namespace std;typedef struct treenode{    struct treenode *l;    struct treenode *r;    int value;}TreeNode,*TreeNodePtr;bool result = true;int number = 0;void IsBst(TreeNodePtr root)//If it is BST then return the number of the nodes it contains else return 0{    if (!root)        return;    static int  lastvalue = INT_MIN;    static int curvalue;    if (!root->l)        IsBst(root->l);    curvalue = root->value;    if (lastvalue >= curvalue)        result = false;    lastvalue = curvalue;    if (!root->r)        IsBst(root->r);}void CountBst(TreeNodePtr root){    if (!root)        return;    CountBst(root->l);    number++;    CountBst(root->r);}int FindTheMaxBst(TreeNodePtr root){    if (!root)        return 0;    result = true;    IsBst(root);    if (result)    {        number = 0;        CountBst(root);        return number;    }    else    {        int lmax = FindTheMaxBst(root->l);        int rmax = FindTheMaxBst(root->r);        return lmax >= rmax ? lmax : rmax;    }}

上面的代码不是o(n)的复杂度

下面的代码对每棵树统计了是不是BST 如果是最大的value 最小value
size是多少 然后将这个信息进行返回是o(n)的复杂度

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public int largestBSTSubtree(TreeNode root) {        int [] res = {0};        helper(root, res);        return res[0];    }    private Node helper(TreeNode root, int [] res){        Node cur = new Node();        if(root == null){            cur.isBST = true;            return cur;        }        Node left = helper(root.left, res);        Node right = helper(root.right, res);        if(left.isBST && root.val > left.max && right.isBST && root.val < right.min){            cur.isBST = true;            cur.min = Math.min(root.val, left.min);            cur.max = Math.max(root.val, right.max);            cur.size = left.size + right.size + 1;            if(cur.size > res[0]){                res[0] = cur.size;            }        }        return cur;    }}class Node{    boolean isBST;    int min;    int max;    int size;    public Node(){        isBST = false;        min = Integer.MAX_VALUE;        max = Integer.MIN_VALUE;        size = 0;    }}