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leetcode 226 Invert Binary Tree 翻转二叉树

2015-06-23 23:53 621 查看
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.


递归解决方案:

/**
* 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;
}
};


非递归解决方案:

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

--------------------------------------------------------------------------------

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非递归解决方案:

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
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