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 boolean isSymmetric(TreeNode root) { if(root == null) return true; return isSymmetric(root.left,root.right); } public boolean isSymmetric(TreeNode left, TreeNode right) { if(left == null) return right == null ; else{ if(right == null) return false; else{ if(left.val != right.val) {return false;} if(!(isSymmetric(left.left,right.right))){ return false;} if(!(isSymmetric(left.right,right.left))) {return false;} } } return true; } }
public class Solution { public boolean isSymmetric(TreeNode root) { if(root == null) return true; if(root.left == null && root.right == null) return true; if(root.left == null || root.right == null) return false; Stack<TreeNode> s1 = new Stack<TreeNode>(); Stack<TreeNode> s2 = new Stack<TreeNode>(); s1.push(root.left); s2.push(root.right); while(!s1.empty()&&!s2.empty()) { TreeNode t1 = s1.pop(); TreeNode t2 = s2.pop(); if(t1 == null && t2 == null) continue; else if(t1 != null && t2 != null && t1.val == t2.val){ s1.push(t1.left); s1.push(t1.right); s2.push(t2.right); s2.push(t2.left); } else{ return false; } } return true; } }