Symmetric Tree

最新推荐文章于 2025-12-30 21:32:24 发布

原创最新推荐文章于 2025-12-30 21:32:24 发布 · 582 阅读

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本文介绍了如何通过递归和迭代方法来判断二叉树是否是对称的，并提供了相应的代码实现。同时，文章还提供了一个错误的解法作为对比，加深理解。

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.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public class Node{
        public int val;
        public int pos;
        public Node(int val,int pos){
            this.val=val;
            this.pos=pos;
        }
    }
   public boolean isSymmetric(TreeNode root) {
        if(root==null) return true;
        else{
            ArrayList<Node> list=new ArrayList<Node>();
            copyBST(root,list,0);
            if(isSym(list)) return true;
            else return false;
        }
    }
    public boolean isSym(List<Node> list){
    	int n=list.size();
        for(int i=0;i<n/2;i++){
        	if(list.get(i).val!=list.get(n-i-1).val || (list.get(i).val==list.get(n-i-1).val && list.get(i).pos==list.get(n-i-1).pos )) return false;
        }
        return true;
    }
    public void copyBST(TreeNode root,List<Node> list,int pos){
        if(root==null) return;
        copyBST(root.left,list,1);
        Node node=new Node(root.val,pos);
        list.add(node);
        copyBST(root.right,list,2);
    }
}

这个解法感觉有问题的，但居然ac了，留作记录，下面是正确的递归解法。

/**
 * Definition for binary tree
 * 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 isSym(root.left,root.right);
    }
    public boolean isSym(TreeNode left,TreeNode right){
    	if(left==null && right==null) return true;
    	if(left!=null && right==null) return false;
    	if(left==null && right!=null) return false;
    	if(left.val!=right.val) return false;
    	else return isSym(left.right,right.left)&&isSym(left.left,right.right);
    }
    
}