Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
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 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; } }
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); }