Solution1：递归 pre-order / 分治
Time Complexity: O(N) Space Complexity: O(N) 递归缓存
Solution2：Iterative pre-order
Time Complexity: O(N) Space Complexity: O(N)
Solution2：BFS
Time Complexity: O(N) Space Complexity: O(N)
Solution1 Code：
class Solution {
public TreeNode invertTree(TreeNode root) {
if(root == null) return null;
TreeNode left_save = root.left;
root.left = invertTree(root.right);
root.right = invertTree(left_save);
return root;
}
}
Solution2 Code：
class Solution {
public TreeNode invertTree(TreeNode root) {
if(root == null) return null;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while(!stack.isEmpty()) {
TreeNode cur = stack.pop();
TreeNode left_save = cur.left;
cur.left = cur.right;
cur.right = left_save;
if(cur.right != null) stack.push(cur.right);
if(cur.left != null) stack.push(cur.left);
}
return root;
}
}
Solution3 Code：
class Solution {
public TreeNode invertTree(TreeNode root) {
if(root == null) return null;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while(!queue.isEmpty()) {
TreeNode cur = queue.poll();
TreeNode left_save = cur.left;
cur.left = cur.right;
cur.right = left_save;
if(cur.right != null) queue.offer(cur.right);
if(cur.left != null) queue.offer(cur.left);
}
return root;
}
}
