[LeetCode] 101. Symmetric Tree Tree symmetry

[LeetCode] 101. Symmetric Tree Tree symmetry

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.

To judge whether a symmetric binary tree.

Solution 1: Non-recursive traversal by layer, each layer is symmetrical check.

Solution 2: recursion,

Wherein the left and right subtree subtree symmetrical conditions: 1) two nodes values ​​are equal, or are empty. 2) the right subtree left subtree node and right node left symmetrical, left subtree right subtree node and right node left symmetrical

Java：Recursion

/**
 * 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 checkNodes(root.left, root.right);
    }

    private boolean checkNodes(TreeNode left, TreeNode right){
        if(left == null && right == null) return true;
        if(left==null || right== null ) return false;
        if(left.val != right.val) return false;
        return checkNodes(left.left, right.right) && checkNodes(left.right, right.left);
    }
}

Java: Iteration

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;
        LinkedList<TreeNode> q1 = new LinkedList<TreeNode>();
        LinkedList<TreeNode> q2 = new LinkedList<TreeNode>();
        q1.add(root.left);
        q2.add(root.right);
        while(!q1.isEmpty() && !q2.isEmpty()){
            TreeNode n1 = q1.poll();
            TreeNode n2 = q2.poll();

            if(n1.val != n2.val)
                return false;
            if((n1.left == null && n2.right != null) || (n1.left != null && n2.right == null))
                return false;
            if((n1.right == null && n2.left != null) || (n1.right != null && n2.left == null))
                return false;

            if(n1.left != null && n2.right != null){
                q1.add(n1.left);
                q2.add(n2.right);
            }

            if(n1.right != null && n2.left != null){
                q1.add(n1.right);
                q2.add(n2.left);
            }
        }
        return true;
    }
}