Easy
Invert Binary Tree — C++
Full explanation · Time O(n) · Space O(h), O(w)
// Time: O(n)
// Space: O(h)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
// Time: O(n)
// Space: O(w), w is the max number of nodes of the levels.
// BFS solution.
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if (root != nullptr) {
queue<TreeNode*> nodes;
nodes.emplace(root);
while (!nodes.empty()) {
auto node = nodes.front();
nodes.pop();
swap(node->left, node->right);
if (node->left != nullptr) {
nodes.emplace(node->left);
}
if (node->right != nullptr) {
nodes.emplace(node->right);
}
}
}
return root;
}
};
// Time: O(n)
// Space: O(h)
// Stack solution.
class Solution2 {
public:
TreeNode* invertTree(TreeNode* root) {
if (root != nullptr) {
stack<TreeNode*> nodes;
nodes.emplace(root);
while (!nodes.empty()) {
auto node = nodes.top();
nodes.pop();
swap(node->left, node->right);
if (node->left != nullptr) {
nodes.emplace(node->left);
}
if (node->right != nullptr) {
nodes.emplace(node->right);
}
}
}
return root;
}
};
// Time: O(n)
// Space: O(h)
// DFS, Recursive solution.
class Solution3 {
public:
TreeNode* invertTree(TreeNode* root) {
if (root != nullptr) {
swap(root->left, root->right);
invertTree(root->left);
invertTree(root->right);
}
return root;
}
};