# 226. Invert Binary Tree
Invert a binary tree.
Example:
Input:
4
/ \
2 7
/ \ / \
1 3 6 9
Output:
4
/ \
7 2
/ \ / \
9 6 3 1
# Solution
Approach 1: Recursion.
Approach 2: Iteration with BFS.
Approach 3: Iteration with DFS.
# Code (Python)
Approach 1:
def invertTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if not root:
return root
self.invertTree(root.left)
self.invertTree(root.right)
root.left, root.right = root.right, root.left
return root
Approach 3:
def invertTree(self, root):
if not root:
return root
stack = [root]
while stack:
node = stack.pop()
node.left, node.right = node.right, node.left
if node.left:
stack.append(node.left)
if node.right:
stack.append(node.right)
return root
# Code (C++)
Approach 1:
/**
* 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)
{
TreeNode *left = invertTree(root->left);
TreeNode *right = invertTree(root->right);
root->left = right;
root->right = left;
}
return root;
}
};
Approach 2:
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
queue<TreeNode*> q;
if (root) q.push(root);
while (!q.empty())
{
TreeNode *node = q.front();
q.pop();
std::swap(node->left, node->right);
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
return root;
}
};