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[LeetCode] Symmetric Tree 判断对称树


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

[LeetCode] Symmetric Tree 判断对称树_算法

2. 110Balanced Binary Tree

[LeetCode] Symmetric Tree 判断对称树_递归_02

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