给你一棵二叉树的根节点 root ，翻转这棵二叉树，并返回其根节点。

提示：

• 树中节点数目范围在 [0, 100] 内
• -100 <= Node.val <= 100

理解题意，翻转即每个结点的左右子树翻转/对调

题解1 递归——自下而上

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(! root) return nullptr;
        TreeNode* LEFT(nullptr), *RIGHT(nullptr);
        /** 递归思路：最后翻转的树 = 翻转后的左子树 + 翻转后的右子树
        	有左子树，翻转左子树
			有右子树，翻转右子树
			每次翻转的步骤一致，所以符合递归思路
		**/
        if(root->left)
            LEFT = invertTree(root->left);
        if(root->right) 
            RIGHT = invertTree(root->right);
        root->left = RIGHT;
        root->right = LEFT;

        return root;
    }
};

题解2 迭代——自上而下

class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(! root) return nullptr;
        // 换成stack，按自上而下思路，逻辑一致
        queue<TreeNode*> Que;
        Que.push(root);
        while(Que.size()){
            TreeNode* tmp = Que.front();
            Que.pop();
            TreeNode* left = tmp->left;
            TreeNode* right = tmp->right;
            if(left) Que.push(left);
            if(right) Que.push(right);
            tmp->left = right;
            tmp->right = left;
        }
        return root;
    }
};