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：
https://www.cnblogs.com/du001011/p/11229170.html

二叉树的层序遍历

public class TreeNode {
    public int data;
    public TreeNode leftChild;
    public TreeNode rightChild;

    public TreeNode(int data){
        this.data = data;
    }
}

public static void levelOrder(TreeNode root){
    LinkedList<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 {
    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 = new TreeNode(2);
        TreeNode n31 = new TreeNode(3);
        TreeNode n32 = new TreeNode(4);
        TreeNode n33 = new TreeNode(4);
        TreeNode n34 = new 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;
        } else {
            return is(root.left, root.right);
        }
    }

    private boolean is(TreeNode n1, TreeNode n2) {
        if (n1 == null || n2 == null) {
            return n1 == n2;
        } else {
            return n1.val == n2.val && is(n1.left, n2.right) && is(n1.right, n2.left);
        }
    }
}

本文地址：https://blog.csdn.net/sinat_42483341/article/details/107282454