Symmetric Tree
描述
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:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following [1,2,2,null,3,null,3]
is not:
1
/ \
2 2
\ \
3 3
Note: Bonus points if you could solve it both recursively and iteratively.
分析
无
递归版
// Symmetric Tree
// 递归版,时间复杂度O(n),空间复杂度O(logn)
public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) return true;
return isSymmetric(root.left, root.right);
}
private static boolean isSymmetric(TreeNode p, TreeNode q) {
if (p == null && q == null) return true; // 终止条件
if (p == null || q == null) return false; // 终止条件
return p.val == q.val // 三方合并
&& isSymmetric(p.left, q.right)
&& isSymmetric(p.right, q.left);
}
}
迭代版
// Symmetric Tree
// 迭代版,时间复杂度O(n),空间复杂度O(logn)
public class Solution {
public boolean isSymmetric (TreeNode root) {
if (root == null) return true;
Stack<TreeNode> s = new Stack<>();
s.push(root.left);
s.push(root.right);
while (!s.isEmpty()) {
TreeNode p = s.pop ();
TreeNode q = s.pop ();
if (p == null && q == null) continue;
if (p == null || q == null) return false;
if (p.val != q.val) return false;
s.push(p.left);
s.push(q.right);
s.push(p.right);
s.push(q.left);
}
return true;
}
}