98.验证二叉搜索树

自己的代码虽然有点石山但是感觉逻辑没啥问题，一些基础例子也过了，就是访问left的left节点时报访问不到null节点，不是是null就返回null呗，有啥访问不到的道理？算了放弃自己想了，学学人家的学会了也算会了

自己的天才石山代码

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public boolean isValidBST(TreeNode root) {return help(root,root.left,root.right);}public boolean help(TreeNode root, TreeNode left,TreeNode right){if(root==null) return true;if(left==null&&right==null) return true;if((left!=null&&left.val>=root.val)||(right!=null&&right.val<=root.val)) return false;if(left.left==null||left.right==null) return  help(right,right.left,right.right);if(right.left==null||right.right==null) return  help(left,left.left,left.right);return help(left,left.left,left.right)&&help(right,right.left,right.right);}
}

 正确代码的思路：由惊人的观察力观察到搜索树的中序遍历就是递增序列，所以只需要中序遍历然后判断是否为递增序列即可，根据这个思路写了ac代码如下

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public boolean isValidBST(TreeNode root) {List<Integer> list = new ArrayList<>();help(root,list);for(int i=0,j=1;j<list.size();i++,j++){if(list.get(i)>=list.get(j)) return false;}return true;}public void help(TreeNode root,List<Integer> list){if(root==null) return;help(root.left,list);list.add(root.val);help(root.right,list); }
}

 但是还有优化空间，可以直接在遍历的过程中进行比较不需要先存数组再遍历数组,这个写法有点拟人了，注意力惊人。这里的思路就不是中序遍历而是对每个三节点进行比较再递归，而min和max时刻维护着对应子树的最大与最小，防止出现三节点均成立但爷孙辈出现问题。这也是我的写法没有考虑到的

 //在遍历过程中判断是否递增class Solution {public boolean isValidBST(TreeNode root) {return help(root,Long.MIN_VALUE,Long.MAX_VALUE);}public boolean help(TreeNode root,long min,long max){if(root==null) return true;if(root.val<=min||root.val>=max) return false;return help(root.left,min,root.val)&&help(root.right,root.val,max);}
}