https://leetcode.cn/problems/validate-binary-search-tree/description/?envType=study-plan-v2&envId=top-interview-150

方法一：中序遍历

考虑只有两个节点和一个结点的情况，可以头尾各加一个最大最小值，不用特判了，也可以直接特判1和0

由于测试样例里有最大值，所以INT最大值不够用

class Solution {List<Long> list = new ArrayList<>();public void dfs(TreeNode root){if(root==null) return;dfs(root.left);list.add((long)root.val);dfs(root.right);}public boolean isValidBST(TreeNode root) {list.add(Long.MIN_VALUE);dfs(root);list.add(Long.MAX_VALUE);for(int i = 1 ;i < list.size()-1; i++)if(list.get(i-1)>=list.get(i)||list.get(i)>=list.get(i+1))return false;return true;}
}

class Solution {List<Integer> list = new ArrayList<>();public void dfs(TreeNode root){if(root==null) return;dfs(root.left);list.add(root.val);dfs(root.right);}public boolean isValidBST(TreeNode root) {dfs(root);if(list.size()==1) return true;if(list.size()==2){if(list.get(1)>list.get(0)) return true;else return false;}for(int i = 1 ;i < list.size()-1; i++)if(list.get(i-1)>=list.get(i)||list.get(i)>=list.get(i+1))return false;return true;}
}

递归法：

非常精妙的区间法，利用区间进行递归。

class Solution {public boolean dfs(TreeNode root,long left,long right){if(root==null) return true;if(root.val>=right||root.val<=left)return false;return dfs(root.left,left,root.val)&&dfs(root.right,root.val,right);}public boolean isValidBST(TreeNode root) {return dfs(root, Long.MIN_VALUE,Long.MAX_VALUE);}
}