手机摄像头过热，关了好几次……

第二题应该不用建树吧

/** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param preStr string字符串 先序遍历序列 * @param midStr string字符串 中序遍历序列 * @return string字符串 */ function getPostOrderOfTree(preStr, midStr) { // write code here let head = new TreeNode(); head.val = preStr[0]; get(head, preStr, midStr); function get(head, pre, mid) { if (pre.length == 0 || mid.length == 0) return; if (pre.length == 1 || mid.length == 1) return; let m = 0; for (let i = 0; i < mid.length; i++) { if (mid[i] == pre[0]) { m = i; break; } } let midLeft = mid.substring(0, m); let midRight = mid.substring(m + 1); let preLeft = pre.substring(1, midLeft.length + 1); let preRight = pre.substring(midLeft.length + 1); let left = null; let right = null; if (preLeft.length > 0) left = new TreeNode(preLeft[0]); if (preRight.length > 0) right = new TreeNode(preRight[0]); head.left = left; head.right = right; get(left, preLeft, midLeft); get(right, preRight, midRight); } let result = []; bl(head, result); console.log(result.join("")); } function TreeNode(val, left, right) { this.val = val; this.left = left == undefined ? null : left; this.right = right == undefined ? null : right; } function bl(head, arr) { if (head == null) return; bl(head.left, arr); bl(head.right, arr); arr.push(head.val); } module.exports = { getPostOrderOfTree: getPostOrderOfTree }; // getPostOrderOfTree("ACDEFHGB", "DECAHFBG")

记错时间了以为是四点 两点十五分发来短信说已经开始了哭了 慌乱双机位 随便支在背后 能照到屏幕和手 😅

第二题先递归得到树 再后序遍历表示没超时