代码随想录算法训练营DAY15 | 二叉树 (2)

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> levelOrder(TreeNode root) {
        Queue queue = new LinkedList<>();
        List> ans = new ArrayList<>();
        if(root == null){
            return ans;
        }
        queue.offer(root);
        while(!queue.isEmpty()){
            //本层节点数
            int size = queue.size();
            List 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