[LeetCode] 101. Symmetric Tree

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?

对称二叉树。

给你一个二叉树的根节点 root ， 检查它是否轴对称。

题目很简单，跟100题几乎一样，也是用DFS思路做。唯一不同的地方是当判断子树的时候，100题是判断left.left == right.left；而这题是判断left.right == right.left

时间O(n)

空间O(n)

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function (root) {
    if (root === null) {
        return true;
    }
    return helper(root.left, root.right);
};

var helper = function (left, right) {
    if (left === null && right === null) {
        return true;
    }
    if (left === null || right === null) {
        return false;
    }
    if (left.val !== right.val) {
        return false;
    }
    return helper(left.left, right.right) && helper(left.right, right.left);
};

 

Java实现

class Solution {
    public boolean isSymmetric(TreeNode root) {
        // corner case
        if (root == null) {
            return true;
        }
        return helper(root.left, root.right);
    }
    
    private boolean helper(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }
        if (left == null || right == null) {
            return false;
        }
        if (left.val == right.val) {
            return helper(left.left, right.right) && helper(left.right, right.left);
        } else {
            return false;
        }
    }
}

 

LeetCode 题目总结