101. Symmetric Tree

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

一刷
题解：用recursion做思路很自然。

/**
 * Definition for a binary tree node.
 * 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 && left == right) return true;
        if((left!=null && right == null) ||(left==null && right != null)) return false;
        if(left.val != right.val) return false;
        return isSymmetric(left.left, right.right) 
            &&  isSymmetric(left.right, right.left); 
    } 
}

二刷
recursion

/**
 * Definition for a binary tree node.
 * 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);
    }
    
    private 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 isSymmetric(left.left, right.right) &&
                isSymmetric(left.right, right.left);
        }
        else return false;
    }
}