LeetCode:Binary Search Tree相关题目合集

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.

解题关键：不能只比较当前root.val与left.val和right.val满不满足BST的定义，需要在递归中记录上一层根节点的值来检测是否满足BST有效性。

Example:

                    5

                  /    \

                4      6

              /   \

            3     7

这是一个无效的BST，虽然4的左右孩子都满足 3 < 4 < 7，但是7大于4的根节点5.

Solution 1: Recursive Traversal

Time Complexity: O(n) Space Complexity: O(1)

public class Solution {
   
    public boolean isValidBST(TreeNode root) {
        return isValid(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
    }

    private boolean isValid(TreeNode root, int min, int max){
        if(root == null){
            return true;
        }

        if(root.val <= min || root.val >= max){
            return false;
        }

        return isValid(root.left, min, root.val) && isValid(root.right, root.val, max);
    }
}

以上代码有一个Bug，在极端情况下，root.val == Integer.MAX_VALUE or root.val == Integer.MIN_VALUE 以上代码会无法正确判断。
先用笨办法解决这个Bug，让我们看看Solution2.

Solution2: In-order Traversal
Time Complexity: O(n) Space Complexity: O(n)

public class Solution {
    public boolean isValidBST(TreeNode root) {
        List<Integer> in = inorder(root);
        for(int i = 1; i < in.size(); i++){
            if(in.get(i) <= in.get(i - 1)){
                return false;
            }
        }
        return true;
    }

    private List<Integer> inorder(TreeNode root){
        List<Integer> result = new ArrayList<Integer>();
        if(root == null){
            return result;
        }
        List<Integer> left = inorder(root.left);
        List<Integer> right = inorder(root.right);
        result.addAll(left);
        result.add(root.val);
        result.addAll(right);