【算法二十六】108. 将有序数组转换为二叉搜索树 98. 验证二叉搜索树

108. 将有序数组转换为二叉搜索树

递归：

/**
 * 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 TreeNode sortedArrayToBST(int[] nums) {
        return sort(nums,0,nums.length-1);
    }
    
    private TreeNode sort(int[] nums,int left,int right){
        if(left > right){
            return null;
        }
        int mid = left +(right - left)/2;
        TreeNode root = new TreeNode(nums[mid]);
        TreeNode l = sort(nums,left,mid-1);
        TreeNode r = sort(nums,mid+1,right);
        root.left = l;
        root.right = r;
        return root;
    }
}

时间复杂度：O(N)

空间复杂度：O(N)

98. 验证二叉搜索树

递归：

/**
 * 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 isValidBSTCheck(root,Long.MIN_VALUE,Long.MAX_VALUE);
    }
     
     private boolean isValidBSTCheck(TreeNode root,long left,long right){
         if(root == null){
             return true;
         }
         
         long x = root.val;
         //二叉树里不允许有重复值
         if(left >= x || x >= right){
             return false;
         }
         
         if(!isValidBSTCheck(root.left,left,x) || !isValidBSTCheck(root.right,x,right)){
             return false;
         }
         return true;
     }
}

时间复杂度：O(N)

空间复杂度：O(N)

递归总结：昨天总结的不是很全面，递归看三点，终止条件，每一层做什么，给上一层返回什么