101. Symmetric Tree-优快云博客

本文链接：https://blog.youkuaiyun.com/daimingyang123/article/details/78939972

本文介绍了一种判断二叉树是否对称的方法，通过递归和迭代两种方式实现。递归方法简单直观，而迭代方法则利用队列存放需要对比的节点。这两种方法均能有效地解决对称二叉树的问题。

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Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

/ \

2 2

/ \ / \

3 4 4 3

But the following [1,2,2,null,3,null,3] is not:

/ \

2 2

\ \

3 3

Note:

Bonus points if you could solve it both recursively and iteratively.

判断一棵树是否是对称树

递归、迭代都写出来了，基本和答案一致，可以说是很屌了

递归比较好想

    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        return isSymmetric(root.left, root.right);
    }
    
    private boolean isSymmetric(TreeNode t1, TreeNode t2) {
        if (t1 == null && t2 == null) return true;
        if (t1 == null || t2 == null) return false;
        return t1.val == t2.val && isSymmetric(t1.left, t2.right) && isSymmetric(t1.right, t2.left);
    }

迭代把需要对比的节点放在一起

public boolean isSymmetric(TreeNode root) {
    Queue<TreeNode> q = new LinkedList<>();
    q.add(root);
    q.add(root);
    while (!q.isEmpty()) {
        TreeNode t1 = q.poll();
        TreeNode t2 = q.poll();
        if (t1 == null && t2 == null) continue;
        if (t1 == null || t2 == null) return false;
        if (t1.val != t2.val) return false;
        q.add(t1.left);
        q.add(t2.right);
        q.add(t1.right);
        q.add(t2.left);
    }
    return true;
}