翻转二叉树
翻转一棵二叉树。
示例:
输入:
4
/ \
2 7
/ \ / \
1 3 6 9
输出:
4
/ \
7 2
/ \ / \
9 6 3 1
思路:
1. 遍历二叉树中每个元素 (遍历方法: 前序、后序、中序、层级遍历)
2.遍历拿到的元素 将元素的左子树 和 右子树 替换
Java实现
public class _226_翻转二叉树 {
/**
* 层序遍历方法
*/
public TreeNode invertTree(TreeNode root) {
if (root == null) return root;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
TreeNode temp = node.left;
node.left = node.right;
node.right = temp;
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
return root;
}
/**
* 中序遍历方法
*/
public TreeNode invertTree3(TreeNode root) {
if (root == null) return root;
invertTree(root.right);
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
// root.right right left 和 right的值已经被调换过了
invertTree(root.right);
return root;
}
/**
* 后序遍历方法
*/
public TreeNode invertTree2(TreeNode root) {
if (root == null) return root;
invertTree(root.left);
invertTree(root.right);
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
return root;
}
/**
* 前序遍历方法
*/
public TreeNode invertTree1(TreeNode root) {
if (root == null) return root;
invertTree(root.left);
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
// root.left 用left原因是 left 和 right的值已经被调换过了
invertTree(root.left);
return root;
}
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
}
Swift实现
func invertTree(_ root: TreeNode?) -> TreeNode? {
if root == nil { return root }
var queue = [TreeNode]();
queue.append(root!);
while !queue.isEmpty {
let node = queue.popLast()
let temp = node?.left
node?.left = node?.right
node?.right = temp
if node?.left != nil {
queue.append((node?.left)!)
}
if node?.right != nil {
queue.append((node?.right)!)
}
}
return root
}
public class TreeNode {
public var val: Int
public var left: TreeNode?
public var right: TreeNode?
init(_ val: Int) {
self.val = val
self.left = nil
self.right = nil
}
}