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Commit dae2e4b

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update 4 leetcode
1 parent b2e3095 commit dae2e4b

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4 files changed

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4 files changed

+211
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/*
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* @lc app=leetcode id=1022 lang=javascript
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*
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* [1022] Sum of Root To Leaf Binary Numbers
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* function TreeNode(val, left, right) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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*/
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/**
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* 递归遍历,然后使用函数参数来保存值
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* @param {TreeNode} root
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* @return {number}
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*/
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var sumRootToLeaf = function(root) {
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let ans = [];
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function dfsSum(node, currentSum = '') {
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if (!node) {
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return null;
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}
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currentSum += node.val;
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if (node.left === null && node.right === null) {
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return ans.push(currentSum);
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} else {
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dfsSum(node.left, currentSum);
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dfsSum(node.right, currentSum);
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}
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}
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dfsSum(root, '');
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return ans.reduce((acc, cur) => acc + parseInt(cur, 2), 0);
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};
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// @lc code=end

‎leetcode/110.balanced-binary-tree.js

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/*
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* @lc app=leetcode id=110 lang=javascript
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*
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* [110] Balanced Binary Tree
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* function TreeNode(val, left, right) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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*/
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/**
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* 类似于 543 题取最大深度
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* @param {TreeNode} root
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* @return {boolean}
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*/
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var isBalanced = function(root) {
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let unBalanced = false;
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function dfs(node) {
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// 当前节点不存在或者已经判定是不平衡的树了,就停止
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if (!node || unBalanced) {
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return 0;
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}
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// 取出左右深度来比较
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let left = dfs(node.left);
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let right = dfs(node.right);
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// 超过平衡了,标记为不平衡
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if (Math.abs(left - right) > 1) {
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unBalanced = true;
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}
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// 取得当前节点的最深的深度,向上返回就是加一
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return Math.max(left, right) + 1;
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}
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dfs(root);
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return !unBalanced;
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};
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// @lc code=end
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/*
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* @lc app=leetcode id=515 lang=javascript
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*
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* [515] Find Largest Value in Each Tree Row
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* function TreeNode(val, left, right) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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*/
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/**
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* 迭代的方式
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* @param {TreeNode} root
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* @return {number[]}
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*/
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var largestValues = function(root) {
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if (!root) {
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return [];
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}
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let queues = [root]; // 保存用于迭代的队列
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let level = 0; // 保存层级
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let res = []; // 保存结果
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while (queues.length) {
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let len = queues.length;
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// 每一层的迭代
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while (len--) {
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const current = queues.shift();
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if (current) {
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// 设置默认值
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res[level] = res[level] || [Number.MIN_SAFE_INTEGER];
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// 如果当前值比较大,就覆盖填入
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if (res[level][0] < current.val) {
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res[level] = [current.val];
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}
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queues.push(current.left);
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queues.push(current.right);
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}
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}
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// 递进层级
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level++;
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}
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return res;
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};
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// @lc code=end
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/*
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* @lc app=leetcode id=897 lang=javascript
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*
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* [897] Increasing Order Search Tree
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* function TreeNode(val, left, right) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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*/
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/**
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* 类似于链表的处理
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* @param {TreeNode} root
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* @return {TreeNode}
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*/
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var increasingBST = function(root) {
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// 新建一棵树节点哑结点
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let dummyNode = new TreeNode(-1);
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// 保存副本,用于遍历,不影响原来节点
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let rootAns = dummyNode;
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function dfs(node) {
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if (!node) {
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return null;
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}
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dfs(node.left);
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// 在遍历过程中就直接实现节点
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rootAns.right = new TreeNode(node.val);
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rootAns.left = null;
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rootAns = rootAns.right;
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dfs(node.right);
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}
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dfs(root);
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// 源节点的右边就是结果
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return dummyNode.right;
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};
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var increasingBST2 = function(root) {
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// 把中序遍历的结果也就是递增的保存到数组中
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let res = [];
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function dfs(node) {
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if (!node) {
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return null;
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}
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dfs(node.left);
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res.push(node.val);
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dfs(node.right);
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}
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dfs(root);
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let index = 0;
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// 遍历数组来创建二叉树
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function createNode() {
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// 当索引到数组长度时停止
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if (index === res.length) {
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return null;
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}
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let node = new TreeNode(res[index]);
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index++;
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// 单向的,所以只需添加右边节点
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node.right = createNode();
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return node;
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}
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return createNode();
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};
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// @lc code=end

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