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invert binary tree.py
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invert binary tree.py
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'''
Invert a binary tree.
4
/ \
2 7
/ \ / \
1 3 6 9
to
4
/ \
7 2
/ \ / \
9 6 3 1
'''
'''
Method: recursion O(n) time O(logn) space
'''
class Solution(object):
def invertTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if not root:
return root
if root.left and root.right:
key = root.left
root.left = root.right
root.right = key
self.invertTree(root.left)
self.invertTree(root.right)
elif root.left:
root.right = root.left
root.left = None
self.invertTree(root.right)
elif root.right:
root.left = root.right
root.right = None
self.invertTree(root.left)
return root
'''
simplify
'''
class Solution(object):
def invertTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if not root:
return root
root.left = self.invertTree(root.right)
root.right = self.invertTree(root.left)
return root
'''
Method: BFS, queue used
'''
public class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return null;
}
final Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()) {
final TreeNode node = queue.poll();
final TreeNode left = node.left;
node.left = node.right;
node.right = left;
if(node.left != null) {
queue.offer(node.left);
}
if(node.right != null) {
queue.offer(node.right);
}
}
return root;
}
}