public static bool dfs(TreeNode root)
{
if (root == null)
{
return true;
}
// Judge the left tree.
if (!dfs(root.left))
{
return false;
}
// judge the sequence.
if (pre != null && root.val <= pre.val)
{
return false;
}
pre = root;
// Judge the right tree.
if (!dfs(root.right))
{
return false;
}
return true;
}
public bool isValidBST_E(TreeNode root)
{
if (root == null)
{
return true;
}
if (!isValidBST_E(root.left))
{
return false;
}
if (!isFirstNode && root.val <= lastVal)
{
return false;
}
isFirstNode = false;
lastVal = root.val;
if (!isValidBST_E(root.right))
{
return false;
}
return true;
}
public static bool isValidBST_D(TreeNode root)
{
// Just use the inOrder traversal to solve the problem.
return dfs(root);
}
/*
* Reference:
*
* https://github.com/yuzhangcmu/LeetCode/blob/master/tree/IsValidBST_1221_2014.java
*
* Just use the inOrder traversal to solve the problem.
*
* Julia's comment:
* 1. review iterative solution inorder traversal algorithm
* thought process, here is the script written:
*
* Iterative solution, need a stack, and two nodes variables to help tracking.
* Start from root node as cur, pre node is null
* Start a loop,
* inside the loop,
* 1. push all the left nodes into the stack;
* 2. if stack is empty, time to exit the loop;
* 3. pop a node from the stack, since no left nodes, just deal with the current node.
* 4. check BST, a value check: based on assumption, inorder traversal output is increasing order.
* 5. Recover for code reuse, next iteration, set cur to pre;
* 6. go to the right node.
*
* 2. online judge passed:
* 74 / 74 test cases passed.
Status: Accepted
Runtime: 172 ms
*/
public bool isValidBST1(TreeNode root)
{
if (root == null)
return true;
Stack<TreeNode> s = new Stack<TreeNode>();
TreeNode cur = root;
TreeNode pre = null;
while (true)
{
while (cur != null) // step 1
{
s.Push(cur);
cur = cur.left;
}
if (s.Count == 0 ) // step 2
break;
cur = s.Pop(); // step 3
if (pre != null && pre.val >= cur.val) // step 4
return false;
pre = cur; // step 5
cur = cur.right; // step 6
}
return true;
}
public static bool isValidBST_C(TreeNode root)
{
if (root == null)
return true;
if (isValidBST_C(root.left) == false) // left subtree
return false;
if (root.val <= previous) // the current node
return false;
previous = root.val;
if (isValidBST_C(root.right) == false) //the right subtree
return false;
return true;
}
public static bool IsValidBST_B(TreeNode node, long MIN, long MAX)
{
if (node == null)
return true;
int v = node.val;
if (!(v > MIN && v < MAX)) // Range: (MIN, MAX)
return false;
if (!IsValidBST_B(node.left, MIN, v)) // left sub tree BST check: new Range: (MIN, v)
return false;
if (!IsValidBST_B(node.right, v, MAX)) // right sub tree BST check: new Range: (v, MAX)
return false;
return true;
}
/*
* julia's comment:
* 1. change int.MinValue to long.MinValue, int.MaxValue to long.MaxValue
* 2. pass online judge:
* 74 / 74 test cases passed.
Status: Accepted
Runtime: 216 ms
* 3. Some analysis about the binary search tree range: Maximum value and minimum value, thought process please:
* Go over an example, binary search tree:
* A
* / \
* B C
* / \ / \
* D E F G
* Level 1: A, A is in range: [long.MIN, long.MAX] <- tip: all nodes are int, to include int.MAX, int.MIN, use long range
* Level 2:
* B, B is in range: [long.MIN, A.value] <- A, go to left , to B, update upper value.
* C, C is in range: [A.value, long.Max] <- A, go to right, to C, update lower value.
* Level 3:
* D, D is in range: [long.MIN, B.value]
* E, E is in range: [B.value, A. value]
* ...
* How to think recursively?
* Every step, there is a value range to maintain
* 4. Read the alpha-beat pruning articles:
* http://blog.csdn.net/likecool21/article/details/23271621
* http://web.cs.ucla.edu/~rosen/161/notes/alphabeta.html
*
*/
public static bool isValidBST_B(TreeNode root)
{
return IsValidBST_B(root, long.MinValue, long.MaxValue);
}
public static bool IsValidBST(TreeNode node, int MIN, int MAX)
{
if (node == null)
return true;
int v = node.val;
bool rootInRange = v > MIN && v < MAX; // Range: (MIN, MAX)
if (!rootInRange)
return false;
bool l_BST = IsValidBST(node.left, MIN, v); // left sub tree BST check: new Range: (MIN, v)
if (!l_BST)
return false;
bool r_BST = IsValidBST(node.right, v, MAX); // right sub tree BST check: new Range: (v, MAX)
if (!r_BST)
return false;
return true;
}
/*
* Spent over 4 hours to go over solutions, on August 28, August 29, 2015
* A list of solutions tried:
* 1. recursive, time O(n), using beta-alpha pruning?
* 2. recursive, use long Max value / min Value instead of int's to avoid the bug
* 3. use inorder traversal output to help checking BST
* 4. same as 3, value TreeNode variable
* 5. sames as 3,
* 6. same as 3,
* 7. same as 3, but use iterative solution
* 8. Great analysis, brute force solution, time complexity O(n^2) vs beta-alpha pruning time O(n)
*/
/*
* Reference:
* http://fisherlei.blogspot.ca/2013/01/leetcode-validate-binary-search-tree.html
*
* analysis from the above blog:
* 对于每一个子树，限制它的最大，最小值，如果超过则返回false。
对于根节点，最大最小都不限制；
每一层节点，左子树限制最大值小于根，右子树最小值大于根；
但是比较有趣的是，在递归的过程中还得不断带入祖父节点的值。
或者，中序遍历该树，然后扫描一遍看是否递增。
* 红字部分是关键点。不把变量遗传下去的话，没法解决如下例子：
{10,5,15,#,#,6,20}
*
* julia's comment:
* 1. Failed to pass online judge:
* [2147483647] input, false, should be true
*/
public static bool isValidBST(TreeNode root)
{
return IsValidBST(root, int.MinValue, int.MaxValue);
}