leetcode 101.Symmetric Tree-对称二叉树|深度遍历

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

【solution】：

An easy method to solve this problem is recursive algorithm. The process is comparing right node and left node of every node. And call this method recursively.

Once right node != left node, return false. Otherwise, return true.

【code】：

public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        return isSymmetric(root.left, root.right);
    }
    public static boolean isSymmetric(TreeNode left, TreeNode right) {
        if(left == null && right == null) return true;
        if(left == null || right == null) return false;
        if(left.val == right.val) return isSymmetric(left.left, right.right)
        		&& isSymmetric(left.right, right.left);
        return false;
    }
}

192 / 192 test cases passed.   Runtime: 1 ms

finally, welcome criticism!