leetcode--Symmetric Tree

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.

confused what

"{1,#,2,3}"

means?
> read more on how binary tree is serialized on OJ.

分类：二叉树
题意：判断一棵二叉树是否对称

解法1：递归。判断是否左右子树是否相等。具体方法是如果都为空，则返回真，否则判断是否其中一个为空，或者值不相等，则返回假。否则最后递归判断左右子树。

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

public boolean solve(TreeNode left,TreeNode right){
if(left==null&&right==null) return true;//
if((left!=null && right==null) || (left==null && right!=null)
|| (left.val!=right.val)) return false;
return solve(left.left, right.right)&&solve(left.right, right.left);
}
}

解法2：使用栈代替递归。左子树顺序入栈，右子树反向入栈。

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if(root==null) return true;
Stack<TreeNode> leftStack = new Stack<TreeNode>();
Stack<TreeNode> rightStack = new Stack<TreeNode>();
leftStack.add(root.left);
rightStack.add(root.right);
while(leftStack.size()>0&&rightStack.size()>0){
TreeNode left = leftStack.pop();
TreeNode right = rightStack.pop();
if(left!=null&&right==null) return false;
if(left==null&&right!=null) return false;
if(left!=null&&right!=null){
if(left.val!=right.val) return false;
//左右子树入栈方向相反
leftStack.add(left.left);
leftStack.add(left.right);
rightStack.add(right.right);
rightStack.add(right.left);
}
}
return true;
}
}