[LeetCode] Symmetric Tree 判断对称树

港控/mmm° 2022-04-13 21:19 101阅读 0赞

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

For example, this binary tree is symmetric:

1
       / \
      2   2
     / \ / \
    3  4 4  3

But the following is not:

1
       / \
      2   2
       \   \
       3    3

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

判断二叉树是否是平衡树，比如有两个节点n1, n2，我们需要比较n1的左子节点的值和n2的右子节点的值是否相等，同时还要比较n1的右子节点的值和n2的左子结点的值是否相等，以此类推比较完所有的左右两个节点。我们可以用递归和迭代两种方法来实现，写法不同，但是算法核心都一样。

public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        return isSameTree(root.left, root.right);
    }
    
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null || q == null) {
            return p == q;
        }
        boolean f = p.val == q.val;
        boolean f2 = isSameTree(p.left, q.right);
        boolean f3 = isSameTree(p.right, q.left);
        return f && f2 && f3;
    }

类似问题:

1. 222Count Complete Tree

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2. 110Balanced Binary Tree

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