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
java
/**
* 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
java
/**
* 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
java
//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);
}
}