翻转一棵二叉树。示例：输入：     4   /   \  2     7 / \   / \1   3 6   9输出：     4   /   \  7     2 / \   / \9   6 3   1备注:这个问题是受到 Max Howell 的 原问题 启发的 ：谷歌：我们90％的工程师使用您编写的软件(Homebrew)，但是您却无法在面试时在白板上写出翻转二叉树这道题，这太糟糕了。

/** * Definition for a binary tree node. * function TreeNode(val) { *     this.val = val; *     this.left = this.right = null; * } *//** * @param {TreeNode} root * @return {TreeNode} */var invertTree = function (root) {  if (!root) return root;  // 递归  //   const left = root.left;  //   const right = root.right;  //   root.right = invertTree(left);  //   root.left = invertTree(right);  // 我们用stack来模拟递归  // 本质上递归是利用了执行栈，执行栈也是一种栈  // 其实这里使用队列也是一样的，因为这里顺序不重要  const stack = [root];  let current = null;  while ((current = stack.shift())) {    const left = current.left;    const right = current.right;    current.right = left;    current.left = right;    if (left) {      stack.push(left);    }    if (right) {      stack.push(right);    }  }  return root;};

# Definition for a binary tree node.# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution:    def invertTree(self, root: TreeNode) -> TreeNode:        if not root:            return None        stack = [root]        while stack:            node = stack.pop(0)            node.left, node.right = node.right, node.left            if node.left:                stack.append(node.left)            if node.right:                stack.append(node.right)        return root

/** * 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* invertTree(TreeNode* root) {        if (root == NULL) return root;        auto q = queue<TreeNode*>();        q.push(root);        while (!q.empty()) {            auto n = q.front(); q.pop();            swap(n->left, n->right);            if (n->left != nullptr) {                q.push(n->left);            }            if (n->right != nullptr) {                q.push(n->right);            }        }        return root;    }};