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.
https://oj.leetcode.com/problems/symmetric-tree/
public boolean isSameTree(TreeNode p, TreeNode q) {
if(p==null&&q==null){
return true;
}
if(p==null){
return false;
}
if(q==null){
return false;
}
Stack<TreeNode> stackA= new Stack<TreeNode>();
Stack<TreeNode> stackB= new Stack<TreeNode>();
stackA.push(p);
stackB.push(q);
while(!stackA.empty()&&!stackB.empty()){
TreeNode a= stackA.pop();
TreeNode b= stackB.pop();
if(a.val!=b.val){
return false;
}
if(a.left!=null){
if(b.left==null){
return false;
}
stackA.push(a.left);
stackB.push(b.left);
}else{
if(b.left!=null){
return false;
}
}
if(a.right!=null){
if(b.right==null){
return false;
}
stackA.push(a.right);
stackB.push(b.right);
}else{
if(b.right!=null){
return false;
}
}
}
return true;
}