[编程题]二叉树-网易

[编程题] 二叉树

有一棵二叉树，树上每个点标有权值，权值各不相同，请设计一个算法算出权值最大的叶节点到权值最小的叶节点的距离。二叉树每条边的距离为1，一个节点经过多少条边到达另一个节点为这两个节点之间的距离。

给定二叉树的根节点root，请返回所求距离。

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) { }
};*/

class Tree {
    //先序遍历找到权重最大和最小叶节点
    int res;
    void preorder(TreeNode* root, TreeNode* &small, TreeNode* &big)
    {
        if (!root) return;

        if (root->left == NULL && root->right == NULL)//是叶子节点
        {
            if ((small == NULL) && (big == NULL))
            {
                small = big = root;
            }
            else
            {
                if (root->val>big->val) big = root;
                if (root->val<small->val) small = root;
            }
        }

        preorder(root->left, small, big);
        preorder(root->right, small, big);
    }

    //计算两个节点之间的距离
    int Dis(TreeNode* root, TreeNode* &small, TreeNode* &big)
    {
        //如果root是NULL
        if (root == NULL) return -1;
        //如果root==big 或者 root==small
        if (root == big || root == small)
            return 0;

        int left = Dis(root->left, small, big);
        int right = Dis(root->right, small, big);
        if (left >= 0 && right >= 0)//分别在左边和右边 
        {
            res=left+1+right+1;
            return left + 1 + right + 1;
        }    
        if (left >=0 ) return left+1;
        if (right>=0)  return right+1;
        
        if (left == -1 && right == -1) return -1;
        return -1;
    }

public:
    int getDis(TreeNode* root) {

        TreeNode* small=NULL, *big=NULL;

        preorder(root, small, big);

        Dis(root, small, big);
        
        return res;

    }
};