一、题目描述
原文链接:226. 翻转二叉树
具体描述:
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:
示例 2:
示例 3:
提示:
- 树中节点数目范围在 [0, 100] 内
- -100 <= Node.val <= 100
二、思路分析
这个题目主要是考查的还是二叉数的遍历!
比如说我们层次遍历的时候,不需要存储值,只需要把左右反转一下就可以来啦!
或者我们进行递归深度优先遍历的时候,只需要把左右孩子啊反转一下就可以啦!需要注意的就是前序遍历和后序遍历都可以反转左右孩子就可以来,但是中序遍历会反转两次,所以不容易实现,可以在纸上画画试试!
三、AC代码
深度遍历翻转
/**
* 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 root;
swapChildren(root);
invertTree(root.left);
invertTree(root.right);
return root;
}
public void swapChildren(TreeNode root){
TreeNode tmpNode = root.left;
root.left = root.right;
root.right = tmpNode;
}
}
层次遍历翻转
/**
* 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) {
Queue<TreeNode> queue = new LinkedList();
if (root != null) queue.offer(root);
while (queue.isEmpty() == false){
int len = queue.size();
while (len > 0){
TreeNode tmpNode = queue.poll();
swapChildren(tmpNode);
if (tmpNode.left != null) queue.offer(tmpNode.left);
if (tmpNode.right != null) queue.offer(tmpNode.right);
len--;
}
}
return root;
}
public void swapChildren(TreeNode root){
TreeNode tmpNode = root.left;
root.left = root.right;
root.right = tmpNode;
}
}
四、总结
- 复习二叉数的遍历方式