给你一棵二叉树的根节点 root ,翻转这棵二叉树,并返回其根节点。
示例 1:
输入:root = [4,2,7,1,3,6,9] 输出:[4,7,2,9,6,3,1] 示例 2:
输入:root = [2,1,3] 输出:[2,3,1] 示例 3:
输入:root = [] 输出:[]
提示:
树中节点数目范围在 [0, 100] 内
-100 <= Node.val <= 100
解法一递归法(DFS)
- 从根开始将左右子树进行交换
- 直到节点为null
import java.security.PublicKey;/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
//确定递归函数的参数和返回值
public TreeNode invertTree(TreeNode root) {
//确定函数的递归边界
if(root == null) {
return null;
}
//确定单层的递归逻辑
//交换左右子树
TreeNode tmp = root.right;
root.right = root.left;
root.left = tmp;
//递归交换当前节点的左子树
invertTree(root.left);
//递归交换当前节点的右子树
invertTree(root.right);
//函数返回时表示当前这个节点和它左右的节点都已经交换完了
return root;
}
}
解法二 迭代法(BFS)
定义队列,一层一层遍历
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) {
return null;
}
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
while (!queue.isEmpty()) {
TreeNode tmp = queue.poll();
// 交换tmp节点的左右子树
TreeNode jiaohuan = tmp.left;
tmp.left = tmp.right;
tmp.right = jiaohuan;
//如果当前节点的左子树不为空,则放入队列等待后续
if (tmp.left != null) {
queue.add(tmp.left);
}
//如果当前节点的右子树不为空, 则放入队列等待后续
if(tmp.right != null){
queue.add(tmp.right);
}
}
return root;
}
}