【判断二叉搜索树】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.

题意：判断给定的二叉树是否为二叉搜索树

递归解法：使用中序遍历，通过设定的pre节点值和当前遍历值相比较，若符合二叉搜索树的定义则返回真

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {        private TreeNode pre=null;        public boolean inorder(TreeNode root){        if(root == null) return true;        boolean left = inorder(root.left);        if(pre != null && pre.val>=root.val)        return false;        pre = root;        boolean right = inorder(root.right);        return left && right;    }        public boolean isValidBST(TreeNode root) {        return inorder(root);    }}

解法二：非递归实现中序遍历，并进行pre和当前遍历值的判断

public class Solution {        public boolean inorder(TreeNode root){        if(root == null) return true;        Stack<TreeNode> s = new Stack<TreeNode>();        TreeNode p = root;                while(p!=null || !s.isEmpty()){            while(p!=null){                s.push(p);                p = p.left;            }            if(!s.isEmpty()){                TreeNode t = s.pop();                if(pre!=null && pre.val>=t.val) return false;                pre = t;                p = t.right;            }        }        return true;    }        public boolean isValidBST(TreeNode root) {        return inorder(root);    }}