LeetCode : 98. Validate Binary Search Tree 二叉搜索树的验证

向右看齐 2021-06-11 15:14 420阅读 0赞

**试题**  
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:

Input:
        2
       / \
      1   3
    Output: true
    Example 2:
    
        5
       / \
      1   4
         / \
        3   6
    Output: false

Explanation: The input is: \[5,1,4,null,null,3,6\]. The root node’s value  
is 5 but its right child’s value is 4.

**代码：**  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzE2MjM0NjEz_size_16_color_FFFFFF_t_70]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzE2MjM0NjEz_size_16_color_FFFFFF_t_70 1]  
对二叉搜索树验证，我们不仅要验证左子节点<父节点<右子节点，其实父节点的父节点还为右边提供了一个下界，为左边提供了一个上界。如图中4位置要求是大于5的。

我们要在当前节点把它的left节点和right节点的上届和下届都要提前比较好。不然到下个节点就无法分清到底是来自于左节点还是右节点。主要比较一下内容：  
当前节点大于左节点、小于右节点；  
左节点大于下届，右节点小于上届。

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

/**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public boolean isValidBST(TreeNode root) {
            if(root==null) return true;
            
            Stack<TreeNode> sta = new Stack<>();
            Stack<Integer> up = new Stack<>();
            Stack<Integer> low = new Stack<>();
            sta.push(root);
            up.push(null);
            low.push(null);
            
            while(!sta.isEmpty()){
                TreeNode cur = sta.pop();
                Integer lower = low.pop();
                Integer upper = up.pop();
                if(cur.left!=null){
                    if(cur.left.val>=cur.val) return false;
                    if(lower!=null && cur.left.val<=lower) return false;
                    sta.push(cur.left);
                    low.push(lower);
                    up.push(cur.val);
                }
                if(cur.right!=null){
                    if(cur.right.val<=cur.val) return false;
                    if(upper!=null && cur.right.val>=upper) return false;
                    sta.push(cur.right);
                    low.push(cur.val);
                    up.push(upper);
                }
            }
            return true;
        }
    }

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