LeetCode OJ：Validate Binary Search Tree（合法的二叉搜索树）

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.

一开始是这样做的：

class Solution {
public:
    bool isValidBST(TreeNode* root) {
       return checkValid(root, INT_MIN, INT_MAX);
    }

    bool checkValid(TreeNode * root, int left, int right)
    {
        if(root == NULL)
            return true;
        return(root->val > left && root->val < right && checkValid(root->left, left, root->val) ]
            && checkValid(root->right, root->val, right));
    }
};

然而并不能通过，因为leetCode增加了两个新的测试用例，把INT_MAX以及INT_MIN也囊括进去了。

那么就只好用中序遍历的方法来了(一开始想的是先中序遍历一遍，然后根据遍历的到的值是不是升序排列的来判断这个二叉树是不是BST，但是后来发现传入一个引用的参数可以实现一边递归一边比较)，代码如下：

class Solution{
public:
    bool isValidBST(TreeNode* root) {
        TreeNode * helper = NULL;
        return inOrder(root, helper);
    }

    bool inOrder(TreeNode * root, TreeNode * & prev){//注意这里的是引用
        if(root == NULL)
            return true;
        bool left = inOrder(root->left, prev);
        if(prev != NULL && prev->val >= root->val)
            return false;
        prev = root;
        bool right = inOrder(root->right, prev);
        return left && right;
    }
};

 

当然上面的也可以采用迭代的方式来做:

class Solution {
public:
    bool isValidBST(TreeNode* root) {
        stack<TreeNode *> s;
        TreeNode * pre = NULL;
        if(root == NULL)
            return true;
        while(root || !s.empty()){
            while(root != NULL){
                s.push(root);
                root = root->left;
            }
            
            root = s.top();
            s.pop();
            if(pre != NULL && root->val <= pre->val) return false;

            pre = root;
            root = root->right;
        }
        return true;
    }
};

 

 

 

下面是java版本的代码，首先还是不可行的Integer.MAX_VALUE方法，但是还是贴下吧：

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

那么只有遍历树了，然后来判断, 这里使用非递归的方法来写:

public class Solution {
    public boolean isValidBST(TreeNode root) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        HashMap<TreeNode, Integer> map = new HashMap<TreeNode, Integer>();
        List<Integer> list = new ArrayList<Integer>();
          if(root == null)
            return true;
          stack.push(root);
          while(!stack.isEmpty()){
              TreeNode node = stack.peek();
              while(node.left != null && !map.containsKey(node.left)){
                  stack.push(node.left);
                  map.put(node.left, 1);
                  node = node.left;
              }
              stack.pop();
              list.add(node.val);
              if(node.right != null && !map.containsKey(node.right)){
                  stack.push(node.right);
                  map.put(node.right, 1);
              }
          }
          for(int i = 0; i < list.size() - 1; ++i){
              if(list.get(i) >= list.get(i+1))
                return false;
          }
          return true;
    }
}