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  • 【LeetCode】Symmetric Tree

    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.

    confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.


    OJ's Binary Tree Serialization:

    The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

    Here's an example:

       1
      / 
     2   3
        /
       4
        
         5
    
    The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
     
    /**
     * Definition for binary tree
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public boolean isSymmetric(TreeNode root) {
            if(root!=null){
                return isMir(root.left,root.right);
            }
            return true;
        }
    
        private boolean isMir(TreeNode left, TreeNode right) {
            // TODO Auto-generated method stub
            if(left==null&&right!=null)
                return false;
            if(left!=null&&right==null)
                return false;
            if(left==null&&right==null)
                return true;
            if(left.val!=right.val)
                return false;
            
            return isMir(left.left,right.right)&&isMir(left.right,right.left);
        }
    }
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  • 原文地址:https://www.cnblogs.com/yixianyixian/p/3716972.html
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