题解 | #二叉树的镜像#

 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param pRoot TreeNode类 
     * @return TreeNode类
     */
    TreeNode* Mirror(TreeNode* pRoot) {
        // write code here
        if(!pRoot)
            return pRoot;
        queue<TreeNode*>q;
        q.push(pRoot);
        TreeNode* node;
        while(!q.empty()){
            int n=q.size();
            for(int i=0;i<n;i++){
                node=q.front();
                q.pop();
                TreeNode* left=node->left;
                TreeNode* right=node->right;
                node->left=right;
                node->right=left;
                if(left)
                    q.push(left);
                if(right)
                    q.push(right);
            }
        }
        return pRoot;
    }
};