第56题 Symmetric Tree

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本文介绍了一种检查二叉树是否为中心对称的方法，提供了递归与非递归两种解决方案。递归方法通过比较左右子树实现，非递归方法采用两个栈分别进行前序遍历和后序遍历。

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

But the following is not:

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.

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Solution in Java 1: 递归方法

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

Solution in Java 2: 非递归的方法

/**
 * 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||root.left==null&&root.right==null) return true;
        else if(root.left==null||root.right==null || root.left.val!=root.right.val) return false;
        Stack<TreeNode> preOrder = new Stack<TreeNode>();
        Stack<TreeNode> postOrder = new Stack<TreeNode>();
        TreeNode lroot = root.left, rroot = root.right;
        while(lroot!=null||rroot!=null|| !preOrder.empty() ||!postOrder.empty()){
             while(lroot!=null || rroot!=null){ //at least one of lroot and rroot is not null then enter the loop
                if(rroot==null || lroot==null ||lroot.val!=rroot.val) return false; 
                //neither rroot nor lroot is null, otherwise return false
                preOrder.push(lroot);
                postOrder.push(rroot);
                lroot = lroot.left;
                rroot = rroot.right;
            }
            if(preOrder.size()!=postOrder.size()) return false;
            if(!preOrder.empty()){
                lroot = preOrder.pop();
                rroot = postOrder.pop();
                lroot = lroot.right;
                rroot=rroot.left;
            }
        }
       
        return true;
    }
   
}

从左右根结点出发，设置两种相反的遍历方式：中左右和中右左，遍历中每个点都相同则结果为true。