java - 为什么 Integer.MIN_VALUE 在平衡二叉树上失败？有什么问题？

Question

https://leetcode.com/problems/balanced-binary-tree/

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

public class Solution {
    public boolean isBalanced(TreeNode root) {
        
        int ret = getLevel(root);
        if(ret < 0)
            return false;
        
        return true;
        
    }
    public int getLevel(TreeNode node) {

        if(node == null)
            return 0;

        int l = getLevel(node.left);
        int r = getLevel(node.right);
        if(Math.abs(l - r) > 1)
            return -99;
            
        return Math.max(l + 1, r + 1);
    }
}

此代码已被接受。 但是，如果我用 Integer.MIN_VALUE 替换 -99，我的代码就会失败。错误是什么？

例如

Input: [1,2,null,3,null,4,null,5]

Output: true

Expected: false

Answer 1

您的代码因溢出的整数运算而失败。以下代码片段将演示这一点:

int val = Integer.MIN_VALUE;
System.out.println(val);
val -= 3;
System.out.println(val);

输出:

-2147483648
2147483645

现在考虑一下您的实际代码中发生了什么:

int l = getLevel(node.left);
// l == -2147483648 == Integer.MIN_VALUE assuming your base case is hit
int r = getLevel(node.right);
// assuming positive r, then Math.abs() will return a massively positive number
if (Math.abs(l - r) > 1)
        return -99;

换句话说，上面的 if 语句将在它真正应该触发 false 时触发 true。

解决方案:

如果您将 getLevel() 方法修改为以下内容，您应该可以避免遇到的问题:

public int getLevel(TreeNode node) {
    if(node == null)
        return 0;

    int l = getLevel(node.left);
    int r = getLevel(node.right);
    if ( (l < 0 ^ r < 0) || Math.abs(l - r) > 1) {
        // you can simply return -1 here, since an actual
        // level should never have a negative value
        return -1;
    }
    else {
        return Math.max(l + 1, r + 1);
    }
}