【leetcode每日刷题】101. Symmetric Tree

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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.

使用递归的方法：

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
 }
 
public class num101 {
    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);       
    }
}

使用迭代的方法：

class Solution {
    public boolean isSymmetric(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        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;
    }
}