我的递归条件是否适合计算二叉树高度？ [英] Are my recursion conditions right to compute binary tree height?

问题描述

我试图在你的帮助下知道我的代码是对还是错，因为遗憾的是我无法运行它来检查。

I'm trying to know whether my code is right or wrong with your help, because sadly I can't run it to check.

没有编译错误。我要做的是找到二叉树的高度。当然，树不必平衡。

There are no compile errors. What I'm trying to do is to find the height of a binary tree. The tree does not have to be balanced, of course.

二进制树中的每个节点可以有两个节点作为子节点

public int height(RBNode t) {
  if (t == null)
  return 0;

  int heightLeft = height(t.left);
  int heightRight = height(t.right);

  if (heightLeft > heightRight) {
    return heightLeft + 1;
  } else {
    return (heightRight + 1);
  }
}

您认为递归条件是否合适？我的朋友声称它总会返回0。

Do you think the recursion conditions are right? My friend claims that it will always return 0.

推荐答案

非常紧凑的版本：

public int height(RBNode t) {
    if (t == null) {
        return 0;
    }
    return Math.max(height(t.left), height(t.right)) + 1;
}

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