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

Note:

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

Tree类题目。

##Approach 1: recursive symmetric tree的特点是root的左右两棵子树呈镜像。

两棵树呈镜像的条件是：

1. 两棵树的root有相同的value；
2. 一颗树的左子树与另一棵树的右子树呈镜像。

因此可采用recursive的方法，一层一层地向下判断，再返回结果。

/**
 * 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 isMirror(root.left, root.right);
    }
    
    private boolean isMirror(TreeNode n1, TreeNode n2) {
        if(n1 == null && n2 == null) return true;
        if(n1 == null || n2 == null) return false;
        return (n1.val == n2.val) && (isMirror(n1.left, n2.right)) && (isMirror(n1.right, n2.left));
    }
}

##Approach 2: iterative

/**
 * 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;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root.left);
        queue.add(root.right);
        while(!queue.isEmpty()) {
            TreeNode n1 = queue.poll();
            TreeNode n2 = queue.poll();
            if(n1 == null && n2 == null) continue;
            if(n1 == null || n2 == null) return false;
            if(n1.val != n2.val) return false;
            queue.add(n1.left);
            queue.add(n2.right);
            queue.add(n1.right);
            queue.add(n2.left);
        }
        return true;
    }
}