本文共 1922 字，大约阅读时间需要 6 分钟。

方法一 递归

/** * 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 boolean isSymmetric(TreeNode root) {
           return help(root, root);    }    //q,p是该节点的左右 子节点    public boolean help(TreeNode p, TreeNode q) {
           if(p == null && q == null) {
               return true;        }        if(p == null || q == null) {
               return false;        }        return p.val == q.val && help(p.left, q.right) && help(p.right, q.left);    }}

方法二 迭代

/** * 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 boolean isSymmetric(TreeNode root) {
           return help(root, root);    }    public boolean help(TreeNode p, TreeNode q) {
           Queue
   
     queue = new LinkedList<>();        queue.offer(p);        queue.offer(q);        while(!queue.isEmpty()) {
               TreeNode u = queue.poll();            TreeNode v = queue.poll();            if(u == null && v == null) {
                   continue;//这里不能直接返回 还需要接着判断队列的其他元素            }            if(u == null || v == null) {
                   return false;            }            if(p.val != q.val) {
                   return false;            }            queue.offer(u.left);            queue.offer(v.right);            queue.offer(u.right);            queue.offer(v.left);        }        return true;    }}
   

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