LeetCode 98. Validate Binary Search Tree Add to List

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.

Example 1:

    2   / \  1   3

Binary tree [2,1,3], return true.

Example 2:

    1   / \  2   3

Binary tree [1,2,3], return false.

解题思路：中序遍历该树，记录访问过的最大节点的值，所有将要访问的节点的值应该大于该值，通过对每个当前访问的节点与目前最大值进行比较，就可以得出结果。

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean isValidBST(TreeNode root) {        //中序遍历，然后将遍历的结果进栈，下一个将要进栈的数一定要大于等于栈顶元素        Stack<Integer> stacknum = new Stack<Integer>();        //本来想用number来记录当前访问的最大值，但是会出现正好用MIN_VAIUE作为节点的特例        //int number=java.lang.Integer.MIN_VALUE;          Stack<TreeNode> stacknode=new Stack<TreeNode>();        if(root==null){            return true;        }else if(root.left==null && root.right==null){             return true;        }else{            //stacknode.push(root);            TreeNode p = root;            while(!stacknode.empty() || p!=null){               // p = stacknode.pop();               // p=p.left;                while(p!=null){                                        stacknode.push(p);                    p=p.left;                }                p=stacknode.pop();                if(stacknum.empty()){                    stacknum.push(p.val);                }else{                    if(stacknum.pop()<p.val){                        stacknum.push(p.val);                    }else{                        return false;                    }                }                /*if(number<p.val){                    number=p.val;                }else{                    return false;                }*/                p=p.right;                                                            }            return true;        }            }};