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.

解法1: 递归Recursive

解法2：迭代Iterative

Java:

public class Solution {
public boolean isValidBST(TreeNode root) {
if (root == null) return true;
return valid(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
public boolean valid(TreeNode root, long low, long high) {
if (root == null) return true;
if (root.val <= low || root.val >= high) return false;
return valid(root.left, low, root.val) && valid(root.right, root.val, high);
}
}

Java:

public boolean isValidBST(TreeNode root) {
if(root==null)
return true;

return helper(root, Double.NEGATIVE_INFINITY, Double.POSITIVE_INFINITY);
}

public boolean helper(TreeNode root, double low, double high){

if(root.val<=low || root.val>=high)
return false;

if(root.left!=null && !helper(root.left, low, root.val)){
return false;
}

if(root.right!=null && !helper(root.right, root.val, high)){
return false;
}

return true;
}　　

Java: Iterative

public boolean isValidBST(TreeNode root) {
if (root == null) return true;
Stack<TreeNode> stack = new Stack<>();
TreeNode pre = null;
while (root != null || !stack.isEmpty()) {
while (root != null) {
stack.push(root);
root = root.left;
}
root = stack.pop();
if(pre != null && root.val <= pre.val) return false;
pre = root;
root = root.right;
}
return true;
}　　

Java: Iterative

public class Solution {
public boolean isValidBST(TreeNode root) {
if(root == null)
return true;

LinkedList<BNode> queue = new LinkedList<BNode>();
queue.add(new BNode(root, Double.NEGATIVE_INFINITY, Double.POSITIVE_INFINITY));
while(!queue.isEmpty()){
BNode b = queue.poll();
if(b.n.val <= b.left || b.n.val >=b.right){
return false;
}
if(b.n.left!=null){
queue.offer(new BNode(b.n.left, b.left, b.n.val));
}
if(b.n.right!=null){
queue.offer(new BNode(b.n.right, b.n.val, b.right));
}
}
return true;
}
}
//define a BNode class with TreeNode and it's boundaries
class BNode{
TreeNode n;
double left;
double right;
public BNode(TreeNode n, double left, double right){
this.n = n;
this.left = left;
this.right = right;
}
}　　

Python: wo

class Solution(object):
def isValidBST(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if not root:
return True

return self.helper(root, float('-inf'), float('inf'))

def helper(self, root, mi, mx):
if not root:
return True

if root.val >= mx or root.val <= mi:
return False

return self.helper(root.left, mi, root.val) and self.helper(root.right, root.val, mx)

Python:

# Morris Traversal Solution
class Solution:
# @param root, a tree node
# @return a list of integers
def isValidBST(self, root):
prev, cur = None, root
while cur:
if cur.left is None:
if prev and prev.val >= cur.val:
return False
prev = cur
cur = cur.right
else:
node = cur.left
while node.right and node.right != cur:
node = node.right

if node.right is None:
node.right = cur
cur = cur.left
else:
if prev and prev.val >= cur.val:
return False
node.right = None
prev = cur
cur = cur.right

return True　　

C++:

// Recursion without inorder traversal
class Solution {
public:
bool isValidBST(TreeNode *root) {
return isValidBST(root, LONG_MIN, LONG_MAX);
}
bool isValidBST(TreeNode *root, long mn, long mx) {
if (!root) return true;
if (root->val <= mn || root->val >= mx) return false;
return isValidBST(root->left, mn, root->val) && isValidBST(root->right, root->val, mx);
}
};

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