数据结构与算法面试题二叉树镜像

Posted 2022-01-19 zhiyong_will

题目来源“数据结构与算法面试题80道”。

问题分析：实质上仍然是二叉树的遍历，上述已经介绍过二叉树的不同的遍历方法，当遍历到每一个节点的时候，交换其左右子树，依次下去，直到叶子节点。

方法：

1、递归的方法：

void tree_swap(BSTreeNode *root)
    BSTreeNode *p_tmp = root->m_pLeft;
    root->m_pLeft = root->m_pRight;
    root->m_pRight = p_tmp;


void tree_mirror(BSTreeNode *root)
    if (NULL == root ||
        (NULL == root->m_pLeft && NULL == root->m_pRight))
        return;
    

    tree_swap(root);
    tree_swap(root->m_pLeft);
    tree_swap(root->m_pRight);

2、循环的方法

void tree_mirror_loop(BSTreeNode *root)
    if (NULL == root ||
        (NULL == root->m_pLeft && NULL == root->m_pRight))
        return;
    

    BSTreeNode *p = root;
    queue<BSTreeNode *> q_tree;
    q_tree.push(p);
    while (! q_tree.empty())
        BSTreeNode *q = q_tree.front();
        q_tree.pop();
        if (NULL != q->m_pLeft || NULL != q->m_pRight)
            tree_swap(q);
        
        if (NULL != q->m_pLeft) q_tree.push(q->m_pLeft);
        if (NULL != q->m_pRight) q_tree.push(q->m_pRight);