leetcode每日一题（2020-05-31）：101.对称二叉树

题目描述：给定一个二叉树，检查它是否是镜像对称的。

每日学习：
1.复习二叉树的遍历
2.BFS
3.递归和迭代真的难。。。。

题解1：中序遍历+层次识别

var isSymmetric = function(root) {
    let result = []
    function midOrder(root, k) {
        if(root) {
            midOrder(root.left, k + 1)
            result.push(root.val + '' + k)
            midOrder(root.right, k + 1)
        }
    }
    midOrder(root, 1)
    // let compare = [...result].reverse()
    // let err = 0
    // for(let i = 0; i < result.length; i++) {
    //     if(result[i] != compare[i]) {
    //         err++
    //     }
    // }
    // return err == 0 ? true : false
    let i = 0, j = result.length - 1
    while (i < j) {
        if (result[i] != result[j]) {
            return false;
        }
        i++;
        j--;
    }
    return true
};

题解2：递归

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

题解3：迭代

var isSymmetric = function(root) {
    if(!root) return true
    let stack = [root.left, root.right]
    while(stack.length) {
        let right = stack.pop()
        let left = stack.pop()
        if(left && right) {
            if(left.val !== right.val) return false
            stack.push(left.left)
            stack.push(right.right)
            stack.push(left.right)
            stack.push(right.left)
        } else if(left || right) {
            return false
        }
    }
    return true
};