子树/最近公共祖先

判断一颗二叉树是是否是另一颗树的子树

class Solution {public:     bool test(TreeNode* T1,TreeNode* T2)     {        if(T1==NULL && T2==NULL)            return true;        if(T1==NULL||T2==NULL||T1->val!=T2->val)            return false;        return test(T1->left,T2->left)&&test(T1->right,T2->right);     }    bool isSubtree(TreeNode *T1, TreeNode *T2) {        // write your code here        if( T2==NULL )            return true;        if(T1==NULL)            return false;        if(T1->val==T2->val)        {            if(test(T1,T2))                return true;        }        return isSubtree(T1->left,T2)||isSubtree(T1->right,T2);    }};

判断一棵树是否是完全二叉树

bool istree(TreeNode* root){    queue<TreeNode*> q;    q.push(root);    int count = 1;    int isbintree = 1;    while (!q.empty())    {        count = q.size();        while (count--)        {            Node* tmp = q.front();            q.pop();            cout << tmp->val << endl;            if (tmp->left)            {                if (isbintree == 0)                    return false;                q.push(tmp->left);            }            else                isbintree = 0;            if (tmp->right)            {                if (isbintree == 0)                    return false;                q.push(tmp->right);            }            else                isbintree = 0;        }    }    return true;}

二叉树中两个节点的最近公共祖先

要求考虑以下三种种情况，给出解决方案，并解决：
1：二叉树每个节点有parent（三叉链）
相当于转换成链表求交点问题
2：二叉树是搜索二叉树。
直接找比他们最大还大的节点
3：就是普通二叉树。（尽可能实现时间复杂度为（N ））
通过路径来求

class Solution {public:    /**     * @param root: The root of the binary search tree.     * @param A and B: two nodes in a Binary.     * @return: Return the least common ancestor(LCA) of the two nodes.     */bool GetNodePath(TreeNode* cur, TreeNode* node, vector<TreeNode*>& l){    if (cur == node)    {        l.push_back(cur);        return true;    }    if(cur==NULL)        return false;    l.push_back(cur);    bool found = false;    if (!found && cur->left)        found = GetNodePath(cur->left, node, l);    if (!found && cur->right)        found = GetNodePath(cur->right, node, l);    if (!found)        l.pop_back();    return found;}    TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *A, TreeNode *B) {        // write your code here        if(root==A||root==B||root==NULL)            return root;        vector<TreeNode*> path1;        GetNodePath(root,A,path1);        vector<TreeNode*> path2;        GetNodePath(root,B,path2);        int size=path1.size()>path2.size()?path2.size():path1.size();        int i=0;        TreeNode* ret=NULL;        for(;i<size;i++)        {            if(path1[i]==path2[i])                ret=path1[i];            else                break;        }            return ret;    }       };