二叉树的镜像

问题描述

假设有一棵树:

             1
            /
           2   3
              /
             4   5

那么镜像将是:

              1
             /
            3   2
           /
          5   4

假设节点是这样的结构:

Assume the nodes are of this structure:

struct node{
      node left;
      node right;
      int value;
}

有人可以为此提出一个算法吗?

Can someone suggest an algorithm for this?

推荐答案

听起来像家庭作业.

看起来很简单.编写一个递归例程，深度优先访问每个节点，并构建左右颠倒的镜像树.

It looks very easy. Write a recursive routine that depth-first visits every node and builds the mirror tree with left and right reversed.

struct node *mirror(struct node *here) {

  if (here == NULL)
     return NULL;
  else {

    struct node *newNode = malloc (sizeof(struct node));

    newNode->value = here->value;
    newNode->left = mirror(here->right);
    newNode->right = mirror(here->left);

    return newNode;
  }
}

这会返回一棵新树 - 其他一些答案会在适当的位置执行此操作.取决于你的作业要求你做什么:)

This returns a new tree - some other answers do this in place. Depends on what your assignment asked you to do :)

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