重刷leetcode-day8-[101. 翻转二叉树]

【题目】 101. 对称二叉树

【题目考察】：二叉树

【目标复杂度】： O(n)

【解法一】：递归法

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
    if (!root) return true;
    const compare = (left, right) => {
        if (!left && right) {
            return false;
        }
        if (left && !right) {
            return false;
        }
        if (!left && !right) {
            return true;
        }
        if (left.val !== right.val) {
            return false;
        }

        const isOuterMirror = compare(left.left, right.right);
        const isInnerMirror = compare(left.right, right.left);

        return isOuterMirror && isInnerMirror;
    }

    return compare(root.left, root.right)

};

【解法二】：迭代法

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
    if (!root) return true;

    const queue = [];
    queue.push(root.left);
    queue.push(root.right);

    while(queue.length) {
        let left = queue.shift();
        let right = queue.shift();
        
        if (!left && !right) {
            continue;
        }
        if (!left && right || !right && left) {
            return false;
        }

        if (left.val !== right.val) {
            return false;
        }

        queue.push(left.left);
        queue.push(right.right);
        queue.push(left.right);
        queue.push(right.left);
    }

    return true

};