leetcode || 98、Validate Binary Search Tree

problem：

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.

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题意：判断一颗二叉树是否为搜索二叉树，注意搜索二叉树的性质：左子树的所有节点小于父节点，

右子树的所有节点大于父节点

thinking：

（1）开始试着用递归写了一个程序，对搜索二叉树的定义曲解了。仅仅比较了左孩子和右孩子。是错的。

class Solution {
public:
    bool isValidBST(TreeNode *root) {
        if(root==NULL)
            return true;
        TreeNode *parent = root;
        TreeNode *left_child=parent->left;
        TreeNode *right_child=parent->right;
        bool r=true;
        bool l=true;
        if(left_child!=NULL)
        {
            if(left_child->val<parent->val)
                l=isValidBST(left_child);
            else
                l=false;
        }
        if(right_child!=NULL)
        {
            if(right_child->val>parent->val)
                r=isValidBST(right_child);
            else
                r=false;
        }
        return l&&r;
    }
};

（2）正确解法是：中序遍历。采用DFS（非递归法）中序遍历二叉树，节点的VAL 应该是一个递增序列。

code：

class Solution {
  public:
      bool isValidBST(TreeNode *root)
      {
        vector<int> ret=inorderTraversal(root);
        int n=ret.size();
        if(n<2)
            return true;
        for(int i=1;i<n;i++)
        {
            if(ret[i]<=ret[i-1])
                return false;
        }
        return true;
      }
  protected:
      vector<int> inorderTraversal(TreeNode *root) {
                 vector<int> ret;
                 stack<TreeNode*> _stack;
                 TreeNode *tmp=root;
                 while(tmp!=NULL || !_stack.empty())
                 {
                     if(tmp!=NULL)
                     {
                         _stack.push(tmp);
                         tmp=tmp->left;
                     }
                     else
                     {
                         tmp=_stack.top();
                         _stack.pop();
                         ret.push_back(tmp->val);
                         tmp=tmp->right;
                     }
                 }
               return ret;
          }
  };