Symmetric Tree

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.

 

首先1. top-down 还是 bottom-up.    选择top-down

     2.  用递归. 是否需要内嵌小递归.   需要. 其实就是判定root的是否一样的小递归.

     3.  逻辑运用小递归. left.left,right.right    left.right,right.left

 

/**
 * 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 true;
         return isSymmetric(root.left,root.right);
    }
    public boolean isSymmetric(TreeNode left, TreeNode right) {

        if(left == null && right == null)
            return true;
        if(left == null || right == null)
            return false;

        if(left.val != right.val)
            return false;
        return isSymmetric(left.left,right.right) && isSymmetric(left.right,right.left);
    }
}