update 4 leetcode

Author: kyriejoshua

leetcode/1022.sum-of-root-to-leaf-binary-numbers.js

@@ -0,0 +1,41 @@
+/*
+ * @lc app=leetcode id=1022 lang=javascript
+ *
+ * [1022] Sum of Root To Leaf Binary Numbers
+ */
+
+// @lc code=start
+/**
+ * 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)
+ * }
+ */
+/**
+ * 递归遍历，然后使用函数参数来保存值
+ * @param {TreeNode} root
+ * @return {number}
+ */
+var sumRootToLeaf = function(root) {
+ let ans = [];
+
+ function dfsSum(node, currentSum = '') {
+ if (!node) {
+ return null;
+ }
+ currentSum += node.val;
+ if (node.left === null && node.right === null) {
+ return ans.push(currentSum);
+ } else {
+ dfsSum(node.left, currentSum);
+ dfsSum(node.right, currentSum);
+ }
+ }
+
+ dfsSum(root, '');
+
+ return ans.reduce((acc, cur) => acc + parseInt(cur, 2), 0);
+};
+// @lc code=end

leetcode/110.balanced-binary-tree.js

@@ -0,0 +1,43 @@
+/*
+ * @lc app=leetcode id=110 lang=javascript
+ *
+ * [110] Balanced Binary Tree
+ */
+
+// @lc code=start
+/**
+ * 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)
+ * }
+ */
+/**
+ * 类似于 543 题取最大深度
+ * @param {TreeNode} root
+ * @return {boolean}
+ */
+var isBalanced = function(root) {
+ let unBalanced = false;
+ function dfs(node) {
+ // 当前节点不存在或者已经判定是不平衡的树了，就停止
+ if (!node || unBalanced) {
+ return 0;
+ }
+ // 取出左右深度来比较
+ let left = dfs(node.left);
+ let right = dfs(node.right);
+ // 超过平衡了，标记为不平衡
+ if (Math.abs(left - right) > 1) {
+ unBalanced = true;
+ }
+ // 取得当前节点的最深的深度，向上返回就是加一
+ return Math.max(left, right) + 1;
+ }
+
+ dfs(root);
+
+ return !unBalanced;
+};
+// @lc code=end

leetcode/515.find-largest-value-in-each-tree-row.js

@@ -0,0 +1,53 @@
+/*
+ * @lc app=leetcode id=515 lang=javascript
+ *
+ * [515] Find Largest Value in Each Tree Row
+ */
+
+// @lc code=start
+/**
+ * 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)
+ * }
+ */
+/**
+ * 迭代的方式
+ * @param {TreeNode} root
+ * @return {number[]}
+ */
+var largestValues = function(root) {
+ if (!root) {
+ return [];
+ }
+ let queues = [root]; // 保存用于迭代的队列
+ let level = 0; // 保存层级
+ let res = []; // 保存结果
+
+ while (queues.length) {
+ let len = queues.length;
+
+ // 每一层的迭代
+ while (len--) {
+ const current = queues.shift();
+ if (current) {
+ // 设置默认值
+ res[level] = res[level] || [Number.MIN_SAFE_INTEGER];
+ // 如果当前值比较大，就覆盖填入
+ if (res[level][0] < current.val) {
+ res[level] = [current.val];
+ }
+ queues.push(current.left);
+ queues.push(current.right);
+ }
+ }
+
+ // 递进层级
+ level++;
+ }
+
+ return res;
+};
+// @lc code=end

leetcode/897.increasing-order-search-tree.js

@@ -0,0 +1,74 @@
+/*
+ * @lc app=leetcode id=897 lang=javascript
+ *
+ * [897] Increasing Order Search Tree
+ */
+
+// @lc code=start
+/**
+ * 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)
+ * }
+ */
+/**
+ * 类似于链表的处理
+ * @param {TreeNode} root
+ * @return {TreeNode}
+ */
+var increasingBST = function(root) {
+ // 新建一棵树节点哑结点
+ let dummyNode = new TreeNode(-1);
+ // 保存副本，用于遍历，不影响原来节点
+ let rootAns = dummyNode;
+
+ function dfs(node) {
+ if (!node) {
+ return null;
+ }
+ dfs(node.left);
+ // 在遍历过程中就直接实现节点
+ rootAns.right = new TreeNode(node.val);
+ rootAns.left = null;
+ rootAns = rootAns.right;
+ dfs(node.right);
+ }
+ dfs(root);
+
+ // 源节点的右边就是结果
+ return dummyNode.right;
+};
+
+var increasingBST2 = function(root) {
+ // 把中序遍历的结果也就是递增的保存到数组中
+ let res = [];
+ function dfs(node) {
+ if (!node) {
+ return null;
+ }
+ dfs(node.left);
+ res.push(node.val);
+ dfs(node.right);
+ }
+ dfs(root);
+
+ let index = 0;
+
+ // 遍历数组来创建二叉树
+ function createNode() {
+ // 当索引到数组长度时停止
+ if (index === res.length) {
+ return null;
+ }
+ let node = new TreeNode(res[index]);
+ index++;
+ // 单向的，所以只需添加右边节点
+ node.right = createNode();
+ return node;
+ }
+
+ return createNode();
+};
+// @lc code=end