[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);
}

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