Leetcode-Validate Binary Search Tree

Leetcode-Validate Binary Search Tree
Description:

 

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.

   

  Confused what"{1,#,2,3}"Means? > Read more on how binary tree is serialized on OJ.

  
  OJ's Binary Tree Serialization:

  The serialization of a binary tree follows a level order traversal, where '# 'signiies a path terminator where no node exists below.

  Here's an example:

     1  / \ 2   3    /   4    \     5

  The above binary tree is serialized"{1,2,3,#,#,4,#,#,5}".

   

  Ideas:

  Because the Binary Tree and the value generated by the central traversal of a binary tree are ordered and necessary, you only need to perform the central traversal of the binary tree, store the value of the traversal node to a list, and compare the values in the list in sequence. If it is ordered, the binary tree is a binary sorting tree; otherwise, it is not.

  Of course, a better way is to use a temp to store the value of the previous node, and then compare it in sequence.

  Code:

   

  public boolean isValidBST(TreeNode root) {boolean flag=true;List
     
      list=new ArrayList
      
       ();     if(root==null) return flag;     Stack
       
        st=new Stack
        
         ();     st.push(root);     TreeNode top=null;     while(!st.empty())     {     top=st.peek();     while(top.left!=null)     {     st.push(top.left);     top=top.left;     }     while(top.right==null)     {     list.add(top.val);     st.pop();     if(!st.empty())     top=st.peek();     else     break;     }     if(!st.empty())     {     list.add(top.val);     st.pop();     st.push(top.right);     }          }     int len=list.size();     int num=list.get(0),temp=0;     for(int i=1;i
         
          temp)     {     flag=false;     break;     }     num=temp;     }return flag;    }
         
        
       
      
     

   

  Result: