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# 98.Validate-Binary-Search-Tree

## 题目描述

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:
2
/ \
1 3
Input: [2,1,3]
Output: true

## 代码

### Approach #1 Divide & Conquer

1. 1.
传递upper和lower, 和当前节点进行比较
Complexity Analysis:
• Time complexis: O(n)
• Space complexity: O(N)
public class Solution {
public boolean isValidBST(TreeNode root) {
if (root == null) return true;
return helper(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
private boolean helper(TreeNode root, long lower, long upper) {
if (root == null) return true;
if (root.val >= upper || root.val <= lower) return false;
boolean isLeftValidBST = helper(root.left, lower, root.val);
boolean isRightValidBST = helper(root.right, root.val, upper);
return isLeftValidBST && isRightValidBST;
}
}

### Approach #2 Iteration

class Solution {
public boolean isValidBST(TreeNode root) {
Integer lower = null, upper = null, val;
update(root, lower, upper);
while (!stack.isEmpty()) {
root = stack.poll();
lower = lowers.poll();
upper = uppers.poll();
if (root == null) continue;
val = root.val;
if (lower != null && val <= lower) return false;
if (upper != null && val >= upper) return false;
update(root.right, val, upper);
update(root.left, lower, val);
}
return true;
}
public void update(TreeNode root, Integer lower, Integer upper) {
}
}

### Approach #3 No-Recursion, Access nodes from small to large

class Solution {
public boolean isValidBST(TreeNode root) {
Stack<TreeNode> stack = new Stack<>();
while (root != null) {
stack.push(root);
root = root.left;
}
TreeNode lastNode = null;
while (!stack.isEmpty()) {
TreeNode node = stack.peek();
if (lastNode != null && lastNode.val >= node.val) {
return false;
}
lastNode = node;
if (node.right == null) {
node = stack.pop();
while (!stack.isEmpty() && stack.peek().right == node) {
node = stack.pop();
}
} else {
node = node.right;
while (node != null) {
stack.push(node);
node = node.left;
}
}
}
return false;
}
}