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.
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經典binary tree的題目。用一個區間,初始化爲(min, max)。依次用它檢查樹中的每一點。如果在區間中,就是合理的。否則,返回false。對區間的更新遵守這樣的原則:如果是左子樹的遞歸,就更新max爲root.val;如果是右子樹遞歸,就更新min爲root.val。當然,也可以對樹進行中序遍歷,然後檢查得到的遍歷結果是不是有序的。但是這樣還需要額外的空間,肯定沒有用區間的方法好,就不再贅述了。
具體代碼如下:
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isValidBSTHelper(TreeNode root, int lowerBound, int upperBound) {
if(root==null) return true;
if(root.val>lowerBound && root.val<upperBound) {
return isValidBSTHelper(root.left, lowerBound, root.val) && isValidBSTHelper(root.right, root.val, upperBound);
}
else return false;
}
public boolean isValidBST(TreeNode root) {
return isValidBSTHelper(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
}
}
Version 2: Construct an array through in-order traversal. Determine whether this array is sorted or not.
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public void isValidBST(TreeNode root, ArrayList<Integer> inorder) {
if(root==null) return;
isValidBST(root.left, inorder);
inorder.add(root.val);
isValidBST(root.right, inorder);
return;
}
public boolean isValidBST(TreeNode root) {
ArrayList<Integer> inorder = new ArrayList<Integer>();
isValidBST(root, inorder);
for(int i=0; i<inorder.size()-1; i++) {
if(inorder.get(i)>=inorder.get(i+1)) return false;
}
return true;
}
}