【剑指offer】【树】27.二叉树的镜像

题目链接：https://leetcode-cn.com/problems/er-cha-shu-de-jing-xiang-lcof/

递归

递归的先序遍历二叉树，交换每个节点的左右子节点，即可生成二叉树的镜像

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mirrorTree(TreeNode* root) {
        if(!root) return nullptr;
        auto tmp = root -> left;
        root -> left = mirrorTree(root -> right);
        root -> right = mirrorTree(tmp);
        return root;
    }
};

辅助栈/队列(BFS)

利用栈或队列遍历树的所有节点，交换每个节点的左右子节点
时间复杂度：O(n)
空间复杂度：O(1)

栈

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mirrorTree(TreeNode* root) {
        stack<TreeNode*> s;
        s.push(root);
        while (!s.empty()) {
            TreeNode* node = s.top();
            s.pop();
            if (node == NULL) {
                continue;
            }
            swap(node->left, node->right);
            s.push(node->left); 
            s.push(node->right);
        }
        return root;
    }
};

队列

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mirrorTree(TreeNode* root) {
        if(!root) return root;
        queue<TreeNode*> q;
        q.push(root);

        while(!q.empty())
        {
            auto tmp = q.front();
            q.pop();

            swap(tmp -> left, tmp -> right);
            if(tmp -> left) q.push(tmp -> left);
            if(tmp -> right) q.push(tmp -> right);
        }

        return root;
    }
};