题解 | #二叉树的镜像#

能够避免递归尽量避免

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    TreeNode* Mirror(TreeNode* pRoot) {
      if (pRoot == nullptr) {
        return nullptr;
      }
      
      std::stack<TreeNode *> stack_;
      stack_.push(pRoot); 
      
      while (!stack_.empty()) {
        TreeNode *node = stack_.top();
        stack_.pop();
        
        TreeNode *tmp = node->left;
        node->left = node->right;
        node->right = tmp;
        
        if (node->left) {
          stack_.push(node->left);
        }
        if (node->right) {
          stack_.push(node->right);
        }
      }
      
      return pRoot;
    }
};

递归：

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) {
      if (pRoot == nullptr) {
        return nullptr;
      }
      
      TreeNode *left = Mirror(pRoot->left);
      TreeNode *right = Mirror(pRoot->right);
      
      pRoot->left = right;
      pRoot->right = left;
      
      return pRoot;
    }
};