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
        TreeNode* cur=pRoot;
        queue<TreeNode*> que;
        TreeNode* res=cur;
        if(pRoot!=nullptr){que.push(cur);}
        while (!que.empty()) {
        TreeNode* cur=que.front();
        cur=swapNode(cur);
        que.pop();
        if(cur->left){que.push(cur->left);}
        if(cur->right){que.push(cur->right);}
        }
        return pRoot;
    }
    TreeNode* swapNode(TreeNode* node){
        if(node==nullptr){return node;}
        TreeNode* tem=node->left;
        node->left=node->right;
        node->right=tem;
        return node;
    }
};