101. Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree 

[1,2,2,3,4,4,3]

 is symmetric:

1
/ \
2   2
/ \ / \
3  4 4  3

But the following 

[1,2,2,null,3,null,3]

 is not:

1
/ \
2   2
\   \
3    3

Note:

Bonus points if you could solve it both recursively and iteratively.
对比left.left, right.right && left.right, right.left，如果val相等，继续递归。注意以下几点：
1、如果root==null，返回true

2、如果left或者right等于null，不能访问其左右节点和val，否则会outstack，所以要先判断left或者right等不等于null

代码如下：

/**
* Definition for a binary tree node.
* 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 helper(root.left, root.right);
}
public boolean helper(TreeNode left, TreeNode right) {
/*if (left == null && right == null) {
return true;
}
if (left == null || right == null) {
return false;
}*/
//above lines can simplify as following
if (left == null || right == null) {
return left == right;
}
if (left.val != right.val) {
return false;
}
return helper(left.left, right.right) && helper(left.right, right.left);
}
}

Non-recursive(use Stack)--460ms:

public boolean isSymmetric(TreeNode root) {
if(root==null) return true;

Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode left, right;
if(root.left!=null){
if(root.right==null) return false;
stack.push(root.left);
stack.push(root.right);
}
else if(root.right!=null){
return false;
}

while(!stack.empty()){
if(stack.size()%2!=0) return false;
right = stack.pop();
left = stack.pop();
if(right.val!=left.val) return false;

if(left.left!=null){
if(right.right==null) return false;
stack.push(left.left);
stack.push(right.right);
}
else if(right.right!=null){
return false;
}

if(left.right!=null){
if(right.left==null) return false;
stack.push(left.right);
stack.push(right.left);
}
else if(right.left!=null){
return false;
}
}

return true;
}