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.
/**
* 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);
}
}原文地址:http://blog.csdn.net/dutsoft/article/details/37736711