Description
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
Example 1:1
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2
/ \
1 3
Input: [2,1,3]
Output: true
Example 2:1
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9 5
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1 4
/ \
3 6
Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.
Solution
DFS Solution1
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20/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isValidBST(TreeNode root) {
return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
public boolean isValidBST(TreeNode root, long minVal, long maxVal) {
if (root == null) return true;
if (root.val >= maxVal || root.val <= minVal) return false;
return isValidBST(root.left, minVal, root.val) && isValidBST(root.right, root.val, maxVal);
}
}
Stack Solution, can be used in other questions like k-th samllest element in BST1
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27/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isValidBST(TreeNode root) {
if (root == null) return true;
Stack<TreeNode> st = new Stack<>();
TreeNode pre = null;
while(root != null || !st.isEmpty()){
while(root != null){
st.push(root);
root = root.left;
}
root = st.pop();
if (pre != null && pre.val >= root.val ) return false;
pre = root;
root = root.right;
}
return true;
}
}