98. Validate Binary Search Tree
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
Example 2:
5
/ \
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
Note: Equal case is not acceptable.
Way 1: 比較直覺,當掌握了inorder travelsa之後,這已透過這個方式檢查是否順序正確。
T: O(N)
class Solution {
public boolean isValidBST(TreeNode root) {
Stack<TreeNode> stack = new Stack<>();
boolean ret = true;
TreeNode prev = null;
while(root != null||!stack.isEmpty()){
while(root != null){
stack.push(root);
root = root.left;
}
root = stack.pop();
if(prev != null && prev.val >= root.val){
ret = false;
break;
}
prev = root;
root = root.right;
}
return ret;
}
}
Way 2 : 使用recursion的方式去判node.left 跟 node.right
T: O(N)
條件除了node.left.val < node.val && node.righ.val > node.val之外,
要額外判斷node.left內的所有子node都要比node.val小,同理node.righ內的所有子node都要比node.val大
所以設計遞迴function時,多了上下限(maxValue, minValue)的概念 對左子樹而言,上限是parent node.val, 下限則是沿用parent node的下限 對右子樹而言,上限是parent node上限, 下限則是parent node.val
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);
}
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