Invert a binary tree.
Example:
Input:
4
/ \
2 7
/ \ / \
1 3 6 9
Output:
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 f*** off.
代码
Approach #1 Recursion DFS
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) return null;
TreeNode right = invertTree(root.right);
TreeNode left = invertTree(root.left);
root.left = right;
root.right = left;
return root;
}
}
Approach #2 Iterative BFS
class Solution {
public TreeNode intervertTree(TreeNode root) {
if (root == null) return null;
Queue<TreeNode> queue = new LinkedList();
queue.add(root);
while (!queue.isEmpty()) {
TreeNode current = queue.poll();
TreeNode tmp = current.left;
current.left = current.right;
current.right = tmp;
if (current.left != null) queue.add(current.left);
if (current.right != null) queue.add(current.right);
}
return root;
}
}