Invert a binary tree.
反转二叉树
基本上二叉树的玩意儿用递归都能做
For example
to
My Solution
(Java) Version 1 Time: 1ms:
简单地递归然后调换左右子树
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return null;
}
root.left = invertTree(root.left);
root.right = invertTree(root.right);
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
return root;
}
}
(Java) Version 2 Time: 0ms (By anjani_sureshbhat):
有递归调用自然就有非递归调用，这就是一个迭代的版本，然而我的迭代写得并不好，还要多看，当然还有一个很重要的就是迭代比递归快
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return null;
}
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
while (!q.isEmpty()) {
TreeNode polled = q.poll();
if (polled.left != null || polled.right != null) {
TreeNode temp = polled.left;
polled.left = polled.right;
polled.right = temp;
if (polled.left != null) q.offer(polled.left);
if (polled.right != null) q.offer(polled.right);
}
}
return root;
}
}
