226. 翻转二叉树
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
解法思路:
1、递归(相当于深度优先遍历)
2、迭代(相当于广度优先遍历)
法一:
/*** 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) {// Recursion// Time: O(n), n 为节点数// Space: O(m), m 为树深度if (root == null) return root;TreeNode left = invertTree(root.left);TreeNode right = invertTree(root.right);root.left = right;root.right = left;return root;}
}
法二:
/*** 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) {// Iterator// Time: O(n), n 为节点数// Space: O(n)if (root == null) return root;// 将二叉树的节点逐层防区队列,然后迭代处理队列中的元素Deque<TreeNode> deque = new ArrayDeque<>();deque.addLast(root);while (!deque.isEmpty()) {// 每次从 队列 中取一个节点,对调它的左右子树TreeNode tmp = deque.removeFirst();TreeNode left = tmp.left;tmp.left = tmp.right;tmp.right = left;// 若左子树不为空,则放入队列中进行后续处理if (tmp.left != null) {deque.addLast(tmp.left);}// 若右子树不为空,则放入队列中进行后续处理if (tmp.right != null) {deque.addLast(tmp.right);}}// 返回处理后的结果return root;}
}