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| 1 | +package com.leetcode.trees; |
| 2 | + |
| 3 | +import static org.junit.Assert.assertTrue; |
| 4 | + |
| 5 | +/** |
| 6 | + * Level: Easy |
| 7 | + * Problem Link: https://leetcode.com/problems/symmetric-tree/ |
| 8 | + * Problem Description: |
| 9 | + * Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). |
| 10 | + * |
| 11 | + * For example, this binary tree [1,2,2,3,4,4,3] is symmetric: |
| 12 | + * |
| 13 | + * 1 |
| 14 | + * / \ |
| 15 | + * 2 2 |
| 16 | + * / \ / \ |
| 17 | + * 3 4 4 3 |
| 18 | + * |
| 19 | + * |
| 20 | + * But the following [1,2,2,null,3,null,3] is not: |
| 21 | + * |
| 22 | + * 1 |
| 23 | + * / \ |
| 24 | + * 2 2 |
| 25 | + * \ \ |
| 26 | + * 3 3 |
| 27 | + * |
| 28 | + * |
| 29 | + * Note: |
| 30 | + * Bonus points if you could solve it both recursively and iteratively. |
| 31 | + * |
| 32 | + * @author rampatra |
| 33 | + * @since 2019年07月25日 |
| 34 | + */ |
| 35 | +public class SymmetricTree { |
| 36 | + |
| 37 | + /** |
| 38 | + * Time Complexity: O(n) Because we traverse the entire input tree once, the total run time is O(n), where n is |
| 39 | + * the total number of nodes in the tree. |
| 40 | + * Space Complexity: O(n) The number of recursive calls is bound by the height of the tree. In the worst case, the |
| 41 | + * tree is linear and the height is in O(n). Therefore, space complexity due to recursive calls on the stack is |
| 42 | + * O(n) in the worst case. |
| 43 | + * Runtime: <a href="https://leetcode.com/submissions/detail/246324484/">0 ms</a>. |
| 44 | + * |
| 45 | + * @param root |
| 46 | + * @return |
| 47 | + */ |
| 48 | + public static boolean isSymmetric(TreeNode root) { |
| 49 | + if (root == null) { |
| 50 | + return true; |
| 51 | + } |
| 52 | + |
| 53 | + return isSymmetric(root.left, root.right); |
| 54 | + } |
| 55 | + |
| 56 | + private static boolean isSymmetric(TreeNode leftRoot, TreeNode rightRoot) { |
| 57 | + if (leftRoot == null && rightRoot == null) { |
| 58 | + return true; |
| 59 | + } else if (leftRoot == null || rightRoot == null) { |
| 60 | + return false; |
| 61 | + } |
| 62 | + |
| 63 | + return isSymmetric(leftRoot.left, rightRoot.right) && isSymmetric(leftRoot.right, rightRoot.left) && leftRoot.val == rightRoot.val; |
| 64 | + } |
| 65 | + |
| 66 | + public static void main(String[] args) { |
| 67 | + TreeNode root = new TreeNode(1); |
| 68 | + root.left = new TreeNode(2); |
| 69 | + root.right = new TreeNode(2); |
| 70 | + root.left.left = new TreeNode(4); |
| 71 | + root.left.right = new TreeNode(3); |
| 72 | + root.right.left = new TreeNode(3); |
| 73 | + root.right.right = new TreeNode(4); |
| 74 | + |
| 75 | + assertTrue(isSymmetric(root)); |
| 76 | + } |
| 77 | +} |
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