面试题27:二叉树的镜像

题目描述

输入一棵二叉树，输出它的镜像

思路：遍历二叉树，左右儿子交换

void MirrorRecursively(BinaryTreeNode *pNode)
{
	if(pNode==nullptr)
		return;
	if(pNode->m_pLeft == nullptr && pNode->m_pRight == nullptr)
		return;
	
	BinaryTreeNode *tmp = pNode->m_pLeft;
	pNode->m_pLeft = pNode->m_pRight;
	pNode->m_pRight = tmp;
	MirrorRecursively(pNode->m_pLeft);
	MirrorRecursively(pNode->m_pRight);
}

leetcode

226. 翻转二叉树
     翻转一棵二叉树

class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(root == nullptr) return root;
        if(root->left==nullptr && root->right == nullptr) return root;
        
        TreeNode *tmp = root->left;
        root->left = root->right;
        root->right = tmp;
        
        if(root->left) 
            root->left = invertTree(root->left);
        if(root->right)
            root->right = invertTree(root->right);
        return root;
    }
};

执行用时 : 0 ms, 在Invert Binary Tree的C++提交中击败了100.00% 的用户
内存消耗 : 9.6 MB, 在Invert Binary Tree的C++提交中击败了5.20% 的用户