Easy

recursive method，要對(duì)稱要注意幾點(diǎn)：

• root.left.val = root.right.val
• root.left的左子樹(shù)要和root.right的右子樹(shù)對(duì)稱
• root.right的左子樹(shù)要和root.left的右子樹(shù)對(duì)稱

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null){
            return true;
        }
        return helper(root.left, root.right);
    }
    
    private boolean helper(TreeNode l, TreeNode r){
        if (l == null && r == null){
            return true;
        }
        if (l == null || r == null){
            return false;
        }
        return l.val == r.val && helper(l.left, r.right) && helper(l.right, r.left);        
    }
}

這道題要保證iterative的也會(huì)寫(xiě)：要注意while循環(huán)里當(dāng)leftTemp == null && rightTemp == null的時(shí)候是continue繼續(xù)檢查下面的，而不是直接返回true.比如下面這個(gè)例子：

image.png

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null){
            return true;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root.left);
        queue.offer(root.right);
        while (!queue.isEmpty()){
            TreeNode leftTemp = queue.poll();
            TreeNode rightTemp = queue.poll();
            if (leftTemp == null && rightTemp == null){
                continue;
            }
            if (leftTemp == null || rightTemp == null){
                return false;
            }
            if (leftTemp.val != rightTemp.val){
                return false;
            }
            queue.offer(leftTemp.left);
            queue.offer(rightTemp.right);
            queue.offer(leftTemp.right);
            queue.offer(rightTemp.left);
        }
        return true;
    }
}