题解 | #判断一棵二叉树是否为搜索二叉树和完全二叉树#

public:
    /**
     * 
     * @param root TreeNode类 the root
     * @return bool布尔型vector
     */
    bool isSearchBiTree(TreeNode* root, int low, int high)
    {
        if(!root) return true;
        if(root->val <= low || root->val >= high) return false;
        return isSearchBiTree(root->left,low,root->val) && isSearchBiTree(root->right, root->val,high);
    }
    bool isSearchBiTree(TreeNode* root)
    {
        return isSearchBiTree(root,INT_MIN, INT_MAX);
    }
    bool isCompleteBiTree(TreeNode* root)
    {
        if(!root) return true;
        queue<TreeNode*> qu;
        qu.push(root);
        TreeNode* cur;
        
        while(qu.front())
        {
            cur = qu.front();
            qu.pop();
            qu.push(cur->left);
            qu.push(cur->right);
        }
        while(!qu.empty()&& qu.front() == nullptr) qu.pop();//如果是完全二叉树，空节点后面应该都是空节点，那么将后面的空节点输出后应该为空
        return qu.empty();
    }
    vector<bool> judgeIt(TreeNode* root) { //参考大佬的Java代码的C++实现
        // write code here
        vector<bool> rst;
        rst.emplace_back(isSearchBiTree(root)); //效率比push_back略高
        rst.emplace_back(isCompleteBiTree(root));
        return rst;
    }