101. Symmetric Tree

难度: Easy

题目描述

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

 

Example 1:

Input: root = [1,2,2,3,4,4,3]
Output: true

Example 2:

Input: root = [1,2,2,null,3,null,3]
Output: false

 

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• -100 <= Node.val <= 100

 

Follow up: Could you solve it both recursively and iteratively?

解题思路

代码实现

解决方案

java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        List<Integer> leftList = new ArrayList<>();
        List<Integer> rightList = new ArrayList<>();
        travelTreeNode(root.left, leftList, true);
        travelTreeNode(root.right, rightList, false);
        if (leftList.equals(rightList)) {
            return true;
        }

        return false;

    }

    public void travelTreeNode(TreeNode root, List<Integer> list, boolean left) {
        if (root == null) {
              list.add(null);
            return;
        }
        list.add(root.val);
        if (left) {
            travelTreeNode(root.left, list,left);

            travelTreeNode(root.right, list,left);
        } else {
            travelTreeNode(root.right, list,left);
            travelTreeNode(root.left, list,left);

        }

    }
}