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Commit 1d11cea

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‎README.md‎

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@@ -65,6 +65,7 @@ This is the solutions collection of my LeetCode submissions, most of them are pr
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| 104 | [Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/) | [JavaScript](./src/maximum-depth-of-binary-tree/res.js) | Easy |
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| 107 | [Binary Tree Level Order Traversal II](https://leetcode.com/problems/binary-tree-level-order-traversal-ii/) | [JavaScript](./src/binary-tree-level-order-traversal-ii/res.js) | Easy |
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| 108 | [Convert Sorted Array to Binary Search Tree](https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/) | [JavaScript](./src/convert-sorted-array-to-binary-search-tree/res.js) | Easy |
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| 110 | [balanced-binary-tree](https://leetcode.com/problems/balanced-binary-tree/) | [TypeScript](./src/balanced-binary-tree/res.ts) | Easy |
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| 111 | [Minimum Depth of Binary Tree](https://leetcode.com/problems/minimum-depth-of-binary-tree/) | [JavaScript](./src/minimum-depth-of-binary-tree/res.js) | Easy |
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| 120 | [Triangle](https://leetcode.com/problems/triangle/) | [JavaScript](./src/triangle/res.js) | Medium |
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| 121 | [Best Time to Buy and Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/) | [JavaScript](./src/best-time-to-buy-and-sell-stock/res.js) | Easy |

‎src/balanced-binary-tree/res.ts‎

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/**
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* Definition for a binary tree node.
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*/
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class TreeNode {
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val: number
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left: TreeNode | null
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right: TreeNode | null
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constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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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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function getTreeLength(root: TreeNode | null): number {
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if (!root) {
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return 0;
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} else {
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return 1 + Math.max(getTreeLength(root.left), getTreeLength(root.right));
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}
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}
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/**
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* 自上而下递归
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*
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* 时间复杂度:O(n^2),其中 n 是二叉树中的节点个数。
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* 最坏情况下,二叉树是满二叉树,需要遍历二叉树中的所有节点,时间复杂度是 O(n)。
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* 对于节点 p,如果它的高度是 d,则 height(p) 最多会被调用 d 次(即遍历到它的每一个祖先节点时)。对于平均的情况,一棵树的高度 h 满足 O(h)=O(logn),因为 d≤h,所以总时间复杂度为 O(nlogn)。对于最坏的情况,二叉树形成链式结构,高度为 O(n),此时总时间复杂度为 O(n^2)。
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*
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* 空间复杂度:O(n),其中 n 是二叉树中的节点个数。空间复杂度主要取决于递归调用的层数,递归调用的层数不会超过 n。
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* @param root
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*/
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function isBalanced(root: TreeNode | null): boolean {
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if (!root || !root.left && !root.right) {
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return true;
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} else {
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return isBalanced(root.left) && isBalanced(root.right) && Math.abs(getTreeLength(root.left) - getTreeLength(root.right)) <= 1;
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}
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};
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function getBalancedLength(root: TreeNode | null): number {
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if (!root) {
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return 0;
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} else {
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const leftLength = getBalancedLength(root.left);
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const rightLength = getBalancedLength(root.right);
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if (leftLength < 0 || rightLength < 0 || Math.abs(leftLength - rightLength) > 1) {
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return -1;
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} else {
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return Math.max(leftLength, rightLength) + 1;
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}
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}
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}
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/**
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* 自底向上的递归
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*
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* 复杂度均为 O(n)
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* @param root
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*/
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function isBalanced2(root: TreeNode | null): boolean {
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if (!root || !root.left && !root.right) {
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return true;
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} else {
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return getBalancedLength(root) >= 0;
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}
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};

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