Validate Binary Search Tree

TAGS: TAG_Divide_and_Conquer TAG_Recursion TAG_Binary_Search_Tree TAG_Binary_Tree TAG_Medium

Question

Problem Statement

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.
  • A single node tree is a BST

Example

An example:

  1.   2
  2.  / \
  3. 1   4
  4.    / \
  5.   3   5

The above binary tree is serialized as {2,1,4,#,#,3,5} (in level order).

题解1 - recursion

按照题中对二叉搜索树所给的定义递归判断，我们从递归的两个步骤出发分析：

  1. 基本条件/终止条件 - 返回值需斟酌。
  2. 递归步/条件递归 - 能使原始问题收敛。

终止条件好确定——当前节点为空，或者不符合二叉搜索树的定义，返回值分别是什么呢？先别急，分析下递归步试试先。递归步的核心步骤为比较当前节点的key和左右子节点的key大小，和定义不符则返回false, 否则递归处理。从这里可以看出在节点为空时应返回true, 由上层的其他条件判断。但需要注意的是这里不仅要考虑根节点与当前的左右子节点，还需要考虑左子树中父节点的最小值和右子树中父节点的最大值。否则程序在[10,5,15,#,#,6,20] 这种 case 误判。

由于不仅需要考虑当前父节点，还需要考虑父节点的父节点… 故递归时需要引入上界和下界值。画图分析可知对于左子树我们需要比较父节点中最小值，对于右子树则是父节点中的最大值。又由于满足二叉搜索树的定义时，左子结点的值一定小于根节点，右子节点的值一定大于根节点，故无需比较所有父节点的值，使用递推即可得上界与下界，这里的实现非常巧妙。

C++ - long long

  1. /**
  2.  * Definition of TreeNode:
  3.  * class TreeNode {
  4.  * public:
  5.  *     int val;
  6.  *     TreeNode *left, *right;
  7.  *     TreeNode(int val) {
  8.  *         this->val = val;
  9.  *         this->left = this->right = NULL;
  10.  *     }
  11.  * }
  12.  */
  13. class Solution {
  14. public:
  15.     /**
  16.      * @param root: The root of binary tree.
  17.      * @return: True if the binary tree is BST, or false
  18.      */
  19.     bool isValidBST(TreeNode *root) {
  20.         if (root == NULL) return true;
  22.         return helper(root, LLONG_MIN, LLONG_MAX);
  23.     }
  25.     bool helper(TreeNode *root, long long lower, long long upper) {
  26.         if (root == NULL) return true;
  28.         if (root->val <= lower || root->val >= upper) return false;
  29.         bool isLeftValidBST = helper(root->left, lower, root->val);
  30.         bool isRightValidBST = helper(root->right, root->val, upper);
  32.         return isLeftValidBST && isRightValidBST;
  33.     }
  34. };

C++ - without long long

  1. /**
  2.  * Definition of TreeNode:
  3.  * class TreeNode {
  4.  * public:
  5.  *     int val;
  6.  *     TreeNode *left, *right;
  7.  *     TreeNode(int val) {
  8.  *         this->val = val;
  9.  *         this->left = this->right = NULL;
  10.  *     }
  11.  * }
  12.  */
  13. class Solution {
  14. public:
  15.     /**
  16.      * @param root: The root of binary tree.
  17.      * @return: True if the binary tree is BST, or false
  18.      */
  19.     bool isValidBST(TreeNode *root) {
  20.         if (root == NULL) return true;
  22.         return helper(root, INT_MIN, INT_MAX);
  23.     }
  25.     bool helper(TreeNode *root, int lower, int upper) {
  26.         if (root == NULL) return true;
  28.         if (root->val <= lower || root->val >= upper) {
  29.             bool right_max = root->val == INT_MAX && root->right == NULL;
  30.             bool left_min = root->val == INT_MIN && root->left == NULL;
  31.             if (!(right_max || left_min)) {
  32.                 return false;
  33.             }
  34.         }
  35.         bool isLeftValidBST = helper(root->left, lower, root->val);
  36.         bool isRightValidBST = helper(root->right, root->val, upper);
  38.         return isLeftValidBST && isRightValidBST;
  39.     }
  40. };

Java

  1. /**
  2.  * Definition of TreeNode:
  3.  * public class TreeNode {
  4.  *     public int val;
  5.  *     public TreeNode left, right;
  6.  *     public TreeNode(int val) {
  7.  *         this.val = val;
  8.  *         this.left = this.right = null;
  9.  *     }
  10.  * }
  11.  */
  12. public class Solution {
  13.     /**
  14.      * @param root: The root of binary tree.
  15.      * @return: True if the binary tree is BST, or false
  16.      */
  17.     public boolean isValidBST(TreeNode root) {
  18.         if (root == null) return true;
  20.         return helper(root, Long.MIN_VALUE, Long.MAX_VALUE);
  21.     }
  23.     private boolean helper(TreeNode root, long lower, long upper) {
  24.         if (root == null) return true;
  25.         // System.out.println("root.val = " + root.val + ", lower = " + lower + ", upper = " + upper);
  26.         // left node value < root node value < right node value
  27.         if (root.val >= upper || root.val <= lower) return false;
  28.         boolean isLeftValidBST = helper(root.left, lower, root.val);
  29.         boolean isRightValidBST = helper(root.right, root.val, upper);
  31.         return isLeftValidBST && isRightValidBST;
  32.     }
  33. }

源码分析

为避免节点中出现整型的最大最小值，引入 long 型进行比较。有些 BST 的定义允许左子结点的值与根节点相同，此时需要更改比较条件为root.val > upper. C++ 中 long 可能与 int 范围相同，故使用 long long. 如果不使用比 int 型更大的类型，那么就需要在相等时多加一些判断。

复杂度分析

递归遍历所有节点，时间复杂度为 O(n), 使用了部分额外空间，空间复杂度为 O(1).

题解2 - iteration

联想到二叉树的中序遍历。

Java

  1. /**
  2.  * Definition for a binary tree node.
  3.  * public class TreeNode {
  4.  *     int val;
  5.  *     TreeNode left;
  6.  *     TreeNode right;
  7.  *     TreeNode(int x) { val = x; }
  8.  * }
  9.  */
  10. public class Solution {
  11.     public boolean isValidBST(TreeNode root) {
  12.         Deque<TreeNode> st = new ArrayDeque<>();
  13.         long pre = Long.MIN_VALUE;
  14.         // inorder traverse
  15.         while (root != null || !st.isEmpty()) {
  16.             if (root != null) {
  17.                 st.push(root);
  18.                 root = root.left;
  19.             }
  20.             else {
  21.                 root = st.pop();
  22.                 if (root.val > pre)
  23.                     pre = root.val;
  24.                 else
  25.                     return false;
  26.                 root = root.right;
  27.             }
  28.         }
  29.         return true;
  30.     }
  31. }

Reference