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.
/** * 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,但竟然ac了,留作记录,以下是正确的递归解法。
/** * 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); } }