LeetCode-Validate Binary Search Tree

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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.

可以先按其本身的定义来解决此题：左子树的所有节点小于此节点，右子树的所有节点都大于此节点。当然还需要两个辅助方法，我称之为树的工具类方法，求最大值和最小值。

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

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

    private TreeNode pre = null;
    public boolean isValidBST(TreeNode root) {
        if (root == null)
        	return true;
        if (!isValidBST(root.left)) 
        	return false;
        if (prev != null && prev.val >= root.val) return false;
        prev = root;
        return isValidBST(root.right);
    }

在遍历过程中可以进行很多改造，也就是改造遍历可以得出很多高效的算法！！！最后一种方法值得回味。

LeetCode-Validate Binary Search Tree

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原文地址：http://blog.csdn.net/my_jobs/article/details/47666909