翻转一棵二叉树。
4
/ \
2 7
/ \ / \
1 3 6 9
转换为:
7 2
/ \ / \
9 6 3 1
详见:https://leetcode.com/problems/invert-binary-tree/
Java实现:
递归实现:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if(root==null){
return null;
}
TreeNode node=root.left;
root.left=invertTree(root.right);
root.right=invertTree(node);
return root;
}
}
迭代实现:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if(root==null){
return null;
}
LinkedList<TreeNode> que=new LinkedList<TreeNode>();
que.offer(root);
while(!que.isEmpty()){
TreeNode node=que.poll();
TreeNode tmp=node.left;
node.left=node.right;
node.right=tmp;
if(node.left!=null){
que.offer(node.left);
}
if(node.right!=null){
que.offer(node.right);
}
}
return root;
}
}
C++实现:
方法一:
/**
* 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==nullptr)
{
return nullptr;
}
TreeNode *node=root->left;
root->left=invertTree(root->right);
root->right=invertTree(node);
return root;
}
};
方法二:
/**
* 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==nullptr)
{
return nullptr;
}
queue<TreeNode*> que;
que.push(root);
while(!que.empty())
{
TreeNode *node=que.front();
que.pop();
TreeNode *tmp=node->left;
node->left=node->right;
node->right=tmp;
if(node->left)
{
que.push(node->left);
}
if(node->right)
{
que.push(node->right);
}
}
return root;
}
};