[LeetCode-12]Validate Binary Search Tree

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.

Analysis:

The strigh forwared idea is using the Tree Traversal.  Consider the Binary Search Tree with InOrder traversal. The result should be a sequence in ascending order, otherwise, it's not a binary search tree.

ArrayList<Integer> result;public boolean isValidBST(TreeNode root) {result = new ArrayList<Integer>();if(root == null) return true;inorder(root);for(int i=0;i<result.size()-1;i++){if(result.get(i)>=result.get(i+1))return false;}return true;    }public void inorder(TreeNode root){if(root!=null){inorder(root.left);result.add(root.val);inorder(root.right);}}

the other solution, we could set the bounary min and max value for each sub-tree, 

the root has no restriction

for every level tree, the left child max value < root, the right child min value > root

public boolean isValidBST(TreeNode root) {return judgeBST(root, Integer.MAX_VALUE, Integer.MIN_VALUE);    }public boolean judgeBST(TreeNode root, int max, int min){if(root == null) return true;if(root.val<max && root.val>min && judgeBST(root.left, root.val, min)&& judgeBST(root.right, max, root.val)){return true;}else {return false;}}

c++

bool judgeValidBST(TreeNode *root, int nMin, int nMax){    if(root == NULL) return true;    if(root->val <=nMax && root->val >=nMin       && judgeValidBST(root->left,nMin,root->val)       && judgeValidBST(root->right,root->val,nMax))        return true;    else        return false;}bool isValidBST(TreeNode *root) {    return judgeValidBST(root,INT_MIN, INT_MAX);}