Invert a binary tree.
4
/ \
2 7
/ \ / \
1 3 6 9
to
4
/ \
7 2
/ \ / \
9 6 3 1
Trivia:
This problem was inspired by this original tweet by Max Howell:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.
方法一:递归实现
/**
* 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)
return NULL;
TreeNode* r;
r = root->left;
root->left = invertTree(root->right);
root->right = invertTree(r);
}
};
方法二:利用队列(queue)非递归实现
/**
* 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* Change(TreeNode* needl){
TreeNode* temp;
temp = needl->left;
needl->left = needl->right;
needl->right = temp;
return needl;
}
TreeNode* invertTree(TreeNode* root) {
queue<TreeNode*> BiQueue;
if(!root)
return NULL;
BiQueue.push(root);
TreeNode* needl;
while(!BiQueue.empty()){
needl = BiQueue.front();
BiQueue.pop();
Change(needl);
if(needl->left)
BiQueue.push(needl->left);
if(needl->right)
BiQueue.push(needl->right);
}
return root;
}
};