leetcode 101. 对称二叉树 递归解法

题目

思路

一开始想要层序遍历二叉树，然后判断每一层是否回文，写着写着发现这样太麻烦了，看了评论区别人的递归解法，才写出来的。

二叉树的四种遍历方法：
/d/file/titlepic/11229170.html class TreeNode { public int data; public TreeNode leftChild; public TreeNode rightChild; public TreeNode(int data){ this.data = data; }}public static void levelOrder(TreeNode root){ Li早中晚健康饮食习惯nkedList<TreeNode> queue = new LinkedList<>(); queue.add(root); while(!queue.isEmpty()){ root = queue.pop(); System.out.print(root.data+" "); if(root.leftChild!=null) queue.add(root.leftChild); if(root.rightChild!=null) queue.add(root.rightChild); }}

题解（递归解法）

// Definition for a binary tree node.class TreeNode 边走边看{    int val;    TreeNode left;    TreeNode right;    TreeNode(int x) {        val = x;    }    @Override    public String toString() {        return "" + val;    }}class Solution {    // 测试用例    // 1    // / \    // 2 2    // / \ / \    // 3 4 4 3    public static void main(String[] args) {        Solution solution = new Solution();        TreeNode n1 = new TreeNode(1);        TreeNode n21 = new TreeNode(2);        TreeNode n22淄博驾校模拟考试 = new TreeNode(2邀请信怎么写);        TreeNode n31 = new TreeNode(3);        TreeNode n32 = new TreeNode(4);        TreeNode n33 = new TreeNode(4);        TreeNode n34 = ne英语四级作文万能句型w TreeNode(3);        n1.left = n21;        n1.right = n22;        n21.left = n31;        n21.right = n32;        n22.left = n33;        n22.right = n34;        boolean res = solution.isSymmetric(n1);        System.out.println(res);    }    public boolean isSymmetric(TreeNode root) {        if (root == null) {            return true;        } el {            return is(root.left, root.right);        }    }    private boolean is(TreeNode n1, TreeNode n2) {        if (n1 == null || n2 == null) {            return n1 == n2;        } el {            return n1.val == n2.val && is(n1.left, n2.right) && is(n1.right, n2.left);        }    }}

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