【二叉树是否对称】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 is symmetric:

    1   / \  2   2 / \ / \3  4 4  3

But the following is not:

    1   / \  2   2   \   \   3    3

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

题意：判断是否为对称数

解法一：性能不怎么好的递归，先按原树创建一个镜像再比较两棵树是否一致

若要加强的话，可以直接在原树上比较，递归比较左右节点的值是否相同

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {        public TreeNode mirror(TreeNode root){        if(root == null) return null;        TreeNode left = mirror(root.left);        TreeNode right = mirror(root.right);        TreeNode nr = new TreeNode(root.val);        nr.left = right;        nr.right = left;        return nr;    }        public boolean isSame(TreeNode p1, TreeNode p2){        if(p1==null && p2==null) return true;        if(p1 == null && p2 != null) return false;        if(p1 != null && p2 == null) return false;        return (p1.val == p2.val) && isSame(p1.left, p2.left) && isSame(p1.right, p2.right);    }        public boolean isSymmetric(TreeNode root) {        if(root == null) return true;        TreeNode p = mirror(root);        return isSame(root, p);    }}

解法二：递归，直接比较原树的左右节点值

public class Solution {        public boolean isSame(TreeNode p1, TreeNode p2){        if(p1==null && p2==null) return true;        if(p1 == null && p2 != null) return false;        if(p1 != null && p2 == null) return false;        return (p1.val == p2.val) && isSame(p1.left, p2.right) && isSame(p1.right, p2.left);    }        public boolean isSymmetric(TreeNode root) {        if(root == null) return true;        return isSame(root, root);    }}

解法三：使用两个队列分别遍历树的左右节点，并比较值

public class Solution {    public boolean isSymmetric(TreeNode root) {        if(root == null) return true;        Queue<TreeNode> q1 = new LinkedList<TreeNode>();        Queue<TreeNode> q2 = new LinkedList<TreeNode>();                if(root.left != null) q1.add(root.left);        if(root.right != null) q2.add(root.right);                while(!q1.isEmpty() && !q2.isEmpty()){            TreeNode t1 = q1.remove();            TreeNode t2 = q2.remove();            if(t1.val == t2.val){                if(t1.left != null && t2.right != null){                     q1.add(t1.left);                     q2.add(t2.right);                }//注意判断的条件，只有一个不为空，一个为空才不对称                else if((t1.left!=null && t2.right==null) || (t1.left==null && t2.right!=null))                return false;                if(t1.right != null && t2.left != null){                    q1.add(t1.right);                    q2.add(t2.left);                }                else if((t1.right !=null && t2.left==null) || (t1.right==null && t2.left!=null))                return false;            }            else{                return false;            }        }        if(q1.isEmpty() && q2.isEmpty())        return true;                return false;    }}

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