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.

发现了，tree的题基本都靠recursive

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
       return rec(root.left,root.right);
    }
    public boolean rec(TreeNode p, TreeNode q){
        if(p == null && q == null) return true;//When both left and right are null, meaning end.
        if(p == null || q == null) return false;//Means missing left or right, false.
        return ((p.val == q.val) &&
                 rec(p.left, q.right)&&
                 rec(q.left, p.right));
    }
}

 recursive太费空间，iteration也可以

class Solution {
   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()){
        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;
}
}

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

        Stack<TreeNode> s = new Stack<>();
        s.push(root.left);
        s.push(root.right);

        while (!s.isEmpty()) {
            TreeNode p = s.pop ();
            TreeNode q = s.pop ();

            if (p == null && q == null) continue;
            if (p == null || q == null) return false;
            if (p.val != q.val) return false;

            s.push(p.left);
            s.push(q.right);

            s.push(p.right);
            s.push(q.left);
        }

        return true;
    }
}