一、LeetCode 102 二叉树的层序遍历

题目链接：

102.二叉树的层序遍历https://leetcode.cn/problems/binary-tree-level-order-traversal/

思路：利用队列的先进先出特性，在处理本层节点的同时将下层节点入队，每次处理一层的节点，即可实现层序遍历。

class Solution {public List<List<Integer>> levelOrder(TreeNode root) {Queue<TreeNode> queue = new LinkedList<>();List<List<Integer>> ans = new ArrayList<>();if(root == null){return ans;}queue.offer(root);while(!queue.isEmpty()){//本层节点数int size = queue.size();List<Integer> list = new ArrayList<>();//记录本层节点 并将下层节点入队for(int i = 0; i < size; i++){TreeNode temp = queue.poll();list.add(temp.val);if(temp.left != null){queue.offer(temp.left);}if(temp.right != null){queue.offer(temp.right);}}ans.add(new ArrayList(list));}return ans;}
}
/*** 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;*     }* }*/

 二、LeetCode 226 翻转二叉树

题目链接：

226.翻转二叉树https://leetcode.cn/problems/invert-binary-tree/

思路：利用二叉树递归实现前序遍历的思想，在访问每个节点时进行交换其左右子树的操作

class Solution {public TreeNode invertTree(TreeNode root) {if(root == null){return root;}//翻转左右子树 中TreeNode temp = root.left;root.left = root.right;root.right = temp;//前序递归遍历 左、右invertTree(root.left);invertTree(root.right);return root;}
}
/*** 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;*     }* }*/

 三、LeetCode 101 对称二叉树

题目链接：

101.对称二叉树https://leetcode.cn/problems/symmetric-tree/

思路：利用后序遍历思想，分别判断外层和里层节点的轴对称情况（左子树左右中、右子树右左中），并对各种空节点情况进行处理，从而判断整棵树的对称情况。

 

class Solution {public boolean isSymmetric(TreeNode root) {return judge(root.left,root.right);}public boolean judge(TreeNode left, TreeNode right){//处理空节点及镜像节点值不相等的情况if(left == null && right == null){return true;}else if(left == null && right != null){return false;}else if(left != null && right == null){return false;}else if(left.val != right.val){return false;}//分别判断外层和里层的对称情况boolean out_flag = judge(left.left, right.right);boolean in_flag = judge(left.right, right.left);//只能采用后序遍历，因为需要先判断外层和里层的轴对称情况return out_flag && in_flag;}
}
/*** 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;*     }* }*/

四、今日小结

        今天的二叉树题目只掌握了递归解法，层序遍历相关题目还未刷，需要找时间补上ovo