【leetcode】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.

題解如下：

遞歸版

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        return isMirror(root,root);
    }
    
    public boolean isMirror(TreeNode t1,TreeNode t2) {
        if(t1==null && t2==null)
            return true;
        if(t1 == null || t2 == null)
            return false;
        return (t1.val == t2.val)
            && isMirror(t1.left,t2.right)
            && isMirror(t1.right,t2.left);
    }
}

非遞歸版

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        queue.add(root);
        while(!queue.isEmpty()) {
            TreeNode t1 = queue.poll();
            TreeNode t2 = queue.poll();
            if(t1 == null && t2 == null)
                continue;
            if(t1 == null || t2 == null)
                return false;
            if(t1.val != t2.val)
                return false;
            queue.add(t1.left);
            queue.add(t2.right);
            queue.add(t1.right);
            queue.add(t2.left);
        }
        return true;
    }
}