Solution.java

package Tree.easy.No_101_Symmetric_Tree;


import Tree.TreeNode;
import java.util.LinkedList;
import java.util.Queue;

/**
 * FileName: Solution
 * Author:   EdisonLi的家用MacBook Pro
 * Date:     2019-03-24 04:44
 * Description: 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.
 */

public class Solution {
    public boolean isSymmetric(TreeNode root) {
        return isMirror(root,root);
    }

    //递归判断 左子树 等于 右子数
    private boolean isMirror(TreeNode root1, TreeNode root2) {
        if(root1 == null && root2 == null) return true;
        if (root1 == null || root2 == null) return false;

        return (root1.val == root2.val) && isMirror(root1.left , root2.right) && isMirror(root1.right,root2.left);
    }



    public boolean isSymmetric2(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;
    }
}