https://leetcode.com/problems/symmetric-tree/?envType=study-plan-v2&envId=top-interview-150

/* * Definition for a binary tree node. * class TreeNode { * val: number * left: TreeNode | null * right: TreeNode | null * constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) { * this.val = (val===undefined ? 0 : val) * this.left = (left===undefined ? null : left) * this.right = (right===undefined ? null : right) * } * } */

const leftIter = { [Symbol.iterator]() { const stack = [this] return { next() { const cur = stack.pop() if (cur.right) { stack.push(cur.right) } if (cur.left) { stack.push(cur.left) } let isDone = stack.length === 0 return { done: isDone, value: [cur.left?.val, cur.right?.val] } } } } }

const rightIter = { [Symbol.iterator]() { const stack = [this] return { next() { const cur = stack.pop() if (cur.left) { stack.push(cur.left) } if (cur.right) { stack.push(cur.right) }

let isDone = stack.length === 0 return { done: isDone, value: [cur.right?.val, cur.left?.val] } } } } }

function isSymmetric(root: TreeNode | null): boolean { if (!root) return false let left = root.left let right = root.right if (left?.val !== right?.val) { return false }

if (!left && !right) { return true }

if (left && !right) { return false }

if (!left && right) { return false }

Object.assign(left, leftIter) Object.assign(right, rightIter)

const iterLeft = left[Symbol.iterator]() const iterRight = right[Symbol.iterator]()

let iterLeftObj = iterLeft.next() let iterRightObj = iterRight.next()

while (!iterLeftObj.done) { if (iterLeftObj.value[0] !== iterRightObj.value[0] || iterLeftObj.value[1] !== iterRightObj.value[1]) { return false } iterLeftObj = iterLeft.next() iterRightObj = iterRight.next() }

return iterRightObj.done

};

На собеседование - дают простую задачу Я - здесь нам понадобится 2 итератора Собеседующий: