【LeetCode】Symmetric Tree

发布时间：2020-12-14 05:06:46 所属栏目：大数据来源：网络整理

导读：【Description】 Given a binary tree,check whether it is a mirror of itself (ie,symmetric around its center). For example,this binary tree? [1,2,3,4,3] ?is symmetric: 1 / 2 2 / / 3 4 4 3 ? But the following? [1,null,3] ?is not: 1 / 2 2

【Description】

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

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

    1
   /   2   2
 /  / 3  4 4  3

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

    1
   /   2   2
         3    3

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

【AC code】

Reference:?https://leetcode.com/articles/symmetric-tree/

一、递归　　时间复杂度：O(n)

 1 class Solution {
 2     public boolean isSymmetric(TreeNode root) {
 3         if (root == null) return true;
 4         return isSymmetric(root.left,root.right);
 5     }
 6     private boolean isSymmetric(TreeNode r1,TreeNode r2) {
 7         if (r1 == null && r2 == null) return true;
 8         if ((r1 == null || r2 == null) || r1.val != r2.val) return false;
 9         return isSymmetric(r1.left,r2.right) && isSymmetric(r1.right,r2.left);
10     }
11 }

View Code

二、迭代　　时间复杂度：O(n)

 1 class Solution {
 2     public boolean isSymmetric(TreeNode root) {
 3         Queue<TreeNode> q = new LinkedList<>();
 4         q.add(root);
 5         q.add(root);
 6         while (!q.isEmpty()) {
 7             TreeNode t1 = q.poll();
 8             TreeNode t2 = q.poll();
 9             if (t1 == null && t2 == null) continue;
10             if (t1 == null || t2 == null) return false;
11             if (t1.val != t2.val) return false;
12             q.add(t1.left);
13             q.add(t2.right);
14             q.add(t1.right);
15             q.add(t2.left);
16         }
17         return true;
18     }
19 }

View Code

（编辑：李大同）

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