题目链接:点击打开链接
题目大意:略
解题思路:略
相关企业
- 字节跳动
- 谷歌(Google)
- 亚马逊(Amazon)
- 优步(Uber)
- 微软(Microsoft)
- 苹果(Apple)
AC 代码
- Java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
// 解决方案(1)
class Solution {
public TreeNode invertTree(TreeNode root) {
dfs(root);
return root;
}
private void dfs(TreeNode node) {
if (null == node) {
return;
}
dfs(node.left);
dfs(node.right);
TreeNode t = node.left;
node.left = node.right;
node.right = t;
}
}
// 解决方案(2)
class Solution {
public TreeNode invertTree(TreeNode root) {
if(root == null) return null;
Stack<TreeNode> stack = new Stack<>() {
{ add(root); }};
while(!stack.isEmpty()) {
TreeNode node = stack.pop();
if(node.left != null) stack.add(node.left);
if(node.right != null) stack.add(node.right);
TreeNode tmp = node.left;
node.left = node.right;
node.right = tmp;
}
return root;
}
}
// 解决方案(3)
class Solution {
public TreeNode invertTree(TreeNode root) {
if(root == null) return null;
TreeNode tmp = root.left;
root.left = mirrorTree(root.right);
root.right = mirrorTree(tmp);
return root;
}
}
- C++
/**
* 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) {}
* };
*/
// 解决方案(1)
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if (root == nullptr) return nullptr;
TreeNode* tmp = root->left;
root->left = mirrorTree(root->right);
root->right = mirrorTree(tmp);
return root;
}
};
// 解决方案(2)
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if(root == nullptr) return nullptr;
stack<TreeNode*> stack;
stack.push(root);
while (!stack.empty())
{
TreeNode* node = stack.top();
stack.pop();
if (node->left != nullptr) stack.push(node->left);
if (node->right != nullptr) stack.push(node->right);
TreeNode* tmp = node->left;
node->left = node->right;
node->right = tmp;
}
return root;
}
};