98. 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.

Solution 1:

//相当于中序遍历，先加入中间节点，再加入左节点，再加入右节点

class Solution {

    public boolean isValidBST(TreeNode root) {

        if (root == null)

            return true;

        Stack<TreeNode> s = new Stack<>();

        TreeNode pre = null;

        s.push(root);

        while (!s.isEmpty()) {

            TreeNode tn = s.peek();

            while (tn != null) {

                tn = tn.left;                

                s.push(tn);                            

            }

            s.pop();

            if (!s.isEmpty()) {

                tn = s.pop();

                if (pre != null && tn.val <= pre.val)

                   return false;

                pre = tn;            

                s.push(tn.right);

            }            

        }

        return true;

    }

}

 

Solution 2://必须是Long类型，例如只有有个2^31-1节点，如果把Long改成Integer，则会报错

class Solution {

    Integer prev = null;

    boolean isValid = true;

    public boolean isValidBST(TreeNode root) {

        helper(root);

        return isValid;

    }

    

    private void helper(TreeNode root) {

        if (root == null)

            return;

        if (isValid == false)

            return;

        helper(root.left);

        if (prev != null) {

            if (root.val <= prev)

                isValid = false;

        }

        prev = root.val;

        helper(root.right);               

    }

}

转载于:https://www.cnblogs.com/MarkLeeBYR/p/10537213.html