大牛没有能做出来的题，我们要好好做一做
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.
递归解决方式：
/**
* 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 ==NULL) return root;
TreeNode* node = invertTree(root->left);
root->left = invertTree(root->right);
root->right = node;
return root;
}
};
非递归解决方式：
TreeNode* invertTree(TreeNode* root)
{
if(root == NULL)return NULL;
vector<TreeNode*> stack;
stack.push_back(root);
while(!stack.empty())
{
TreeNode* node = stack.back();// or stack.top()
stack.pop_back();
swap(node->left,node->right);
if(node->left)stack.push_back(node->left);
if(node->right)stack.push_back(node->right);
}
return root;
}
python：
def invertTree(self, root):
if root:
root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)
return root
Maybe make it four lines for better readability:
def invertTree(self, root):
if root:
invert = self.invertTree
root.left, root.right = invert(root.right), invert(root.left)
return root
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And an iterative version using my own stack:
def invertTree(self, root):
stack = [root]
while stack:
node = stack.pop()
if node:
node.left, node.right = node.right, node.left
stack += node.left, node.right
return root
def invertTree(self, root):
if root is None:
return None
root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)
return root
python非递归解决方式：
DFS version:
def invertTree(self, root):
if (root):
self.invertTree(root.left)
self.invertTree(root.right)
root.left, root.right = root.right, root.left
return root
BFS version:
def bfs_invertTree(self, root):
queue = collections.deque()
if (root):
queue.append(root)
while(queue):
node = queue.popleft()
if (node.left):
queue.append(node.left)
if (node.right):
queue.append(node.right)
node.left, node.right = node.right, node.left
return root