LeetCode：101. Symmetric Tree（判断该树是不是对称树）

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.

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方法1：利用递归的方式判断

/**
 * 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 isSymmetry(root, root);    
    }
    public boolean isSymmetry(TreeNode t1, TreeNode t2) {
        if (t1 == null && t2 == null) return true;
        if (t1 == null || t2 == null) return false;
        return (t1.val == t2.val)&& isSymmetry(t1.right, t2.left) && isSymmetry(t1.left, t2.right);
   }
}

时间复杂度：O(n)

空间复杂度：O(n)

方法2：利用队列的方式

public boolean isSymmetric(TreeNode root) {
    Queue<TreeNode> q = new LinkedList<>();
    q.add(root);
    q.add(root);
    while (!q.isEmpty()) {
        TreeNode t1 = q.poll();
        TreeNode t2 = q.poll();
        if (t1 == null && t2 == null) continue;
        if (t1 == null || t2 == null) return false;
        if (t1.val != t2.val) return false;
        q.add(t1.left);
        q.add(t2.right);
        q.add(t1.right);
        q.add(t2.left);
    }
    return true;
}

时间复杂度：O(n)

空间复杂度：O(n)

---

方法3：把方法2进行了改进下，但是在leetcode上面跑，效率都不如方法2；

/**
 * 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 isSymmetry(root,root);
}
    public boolean isSymmetry(TreeNode l1,TreeNode l2){
        if(l1 == null && l2 == null){
            return true;
        }else if(l1 == null || l2==null){
            return false;  
        }else{
            return (l1.val==l2.val)&&(isSymmetry(l1.left,l2.right))&&(isSymmetry(l1.right,l2.left));
        }
    }
}

时间复杂度：O(n)

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空间复杂度:O(n)