题目：

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.

分析：先用中序遍历记下所有节点，然后判断此数组是否是已排序的。

代码如下：

void getroot(TreeNode *root,vector<int> &result)
    {
        if(root->left!=NULL)
        {
            getroot(root->left,result);
        }
        result.push_back(root->val);
        if(root->right!=NULL)
        {
            getroot(root->right,result);
        }
        return;
    }
    bool isValidBST(TreeNode *root) {
        if(root==NULL)return true;
        vector<int> result;
        getroot(root,result);
        for(int i=1;i<result.size();i++)
        {
            if(result[i-1]>=result[i])
            {
                return false;
            }
        }
        return true;
    }

题目：

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

代码如下：

bool isSymmetricTree(TreeNode *p, TreeNode *q) {
        if(p==NULL&&q==NULL)return true;
        if(p==NULL||q==NULL)return false;
        if(p->val!=q->val)
        {
            return false;
        }
        else
        {
            return isSymmetricTree(p->left, q->right)&&isSymmetricTree(p->right, q->left);
        }
    }
    bool isSymmetric(TreeNode *root) {
        if(root==NULL)return true;
        return isSymmetricTree(root->left,root->right);   
    }

题目：

Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

代码如下：

bool isSameTree(TreeNode *p, TreeNode *q) {
        if(p==NULL&&q==NULL)return true;
        if(p==NULL||q==NULL)return false;
        if(p->val!=q->val)
        {
            return false;
        }
        else
        {
            return isSameTree(p->left, q->left)&&isSameTree(p->right, q->right);
        }
    }