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
     */
  	//观察发现只需要 每棵树交换他的子树就可以完成任务 
    //交换子树的左右结点  
    void SwapChildNode(TreeNode* tree)
    {
        if(tree == nullptr) return;
        //交换左右子树
        TreeNode* p = tree->left;
        tree->left = tree->right;
        tree->right = p;
        //进入下一层  
        SwapChildNode(tree->left);
        SwapChildNode(tree->right);
    }

    TreeNode* Mirror(TreeNode* pRoot) {
        // write code here
        if(pRoot == nullptr) return nullptr;
        SwapChildNode(pRoot);
        return pRoot;
    }
};