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;
    }
}