add js solution for leetcode 94 144 145

Author: swiftwind0405

README.md

@@ -281,7 +281,7 @@
 |148|[Sort List](https://leetcode.com/problems/sort-list/)| |Medium|
 |147|[Insertion Sort List](https://leetcode.com/problems/insertion-sort-list/)| |Medium|
 |146|[LRU Cache](https://leetcode.com/problems/lru-cache/)| |Hard|
-|145|[Binary Tree Postorder Traversal](https://leetcode.com/problems/binary-tree-postorder-traversal/)| |Hard|
+|145|[Binary Tree Postorder Traversal](https://leetcode.com/problems/binary-tree-postorder-traversal/)| [js](./algorithms/binaryTreePostorderTraversal/binaryTreePostorderTraversal.js) |Hard|
 |144|[Binary Tree Preorder Traversal](https://leetcode.com/problems/binary-tree-preorder-traversal/)| [js](./algorithms/binaryTreePreorderTraversal/binaryTreePreorderTraversal.js) |Medium|
 |143|[Reorder List](https://leetcode.com/problems/reorder-list/)| |Medium|
 |142|[Linked List Cycle II](https://leetcode.com/problems/linked-list-cycle-ii/)| |Medium|
@@ -332,7 +332,7 @@
 |97|[Interleaving String](https://leetcode.com/problems/interleaving-string/)| |Hard|
 |96|[Unique Binary Search Trees](https://leetcode.com/problems/unique-binary-search-trees/)| |Medium|
 |95|[Unique Binary Search Trees II](https://leetcode.com/problems/unique-binary-search-trees-ii/)| |Medium|
-|94|[Binary Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal/)| |Medium|
+|94|[Binary Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal/)| [js](./algorithms/binaryTreeInorderTraversal/binaryTreeInorderTraversal.js) |Medium|
 |93|[Restore IP Addresses](https://leetcode.com/problems/restore-ip-addresses/)| |Medium|
 |92|[Reverse Linked List II](https://leetcode.com/problems/reverse-linked-list-ii/)| |Medium|
 |91|[Decode Ways](https://leetcode.com/problems/decode-ways/)| |Medium|

algorithms/binaryTreeInorderTraversal/binaryTreeInorderTraversal.js

@@ -0,0 +1,28 @@
+var inorderTraversal = function(root) {
+    if (!root) return [];
+
+    const res = [];
+    let cur1 = root;
+    let cur2;
+
+    while(cur1) {
+        cur2 = cur1.left;
+        //构建连接线
+        if(cur2) {
+            while (cur2.right && cur2.right !== cur1) {
+                cur2 = cur2.right;
+            }
+            if (!cur2.right) {
+                cur2.right = cur1;
+                cur1 = cur1.left;
+                continue;
+            } else {
+                cur2.right = null;
+            }
+        }
+        res.push(cur1.val);
+        cur1 = cur1.right;
+    }
+    
+    return res;
+};

algorithms/binaryTreePostorderTraversal/binaryTreePostorderTraversal.js

@@ -0,0 +1,52 @@
+var postorderTraversal = function(root) {
+    if (!root) return [];
+
+    const res = [];
+
+    // 翻转单链表
+    const postMorrisReverseList = (root) => {
+        let cur = root;
+        let pre;
+        while (cur) {
+            let next = cur.right;
+            cur.right = pre;
+            pre = cur;
+            cur = next;
+        }
+	    return pre;
+    };
+    // 打印函数
+    const postMorrisPrint = (root) => {
+        const reverseList = postMorrisReverseList(root);
+        let cur = reverseList;
+        while (cur) {
+            res.push(cur.val);
+            cur = cur.right;
+        }
+        postMorrisReverseList(reverseList);
+    };
+
+    let cur1 = root;    // 遍历树的指针变量
+    let cur2;           // 当前子树的最右节点
+
+    while(cur1) {
+        cur2 = cur1.left;
+        if(cur2) {
+            while (cur2.right && cur2.right !== cur1) {
+                cur2 = cur2.right;
+            }
+            if (!cur2.right) {
+                cur2.right = cur1;
+                cur1 = cur1.left;
+                continue;
+            } else {
+                cur2.right = null;
+                postMorrisPrint(cur1.left);
+            }
+        }
+        cur1 = cur1.right;
+    }
+    postMorrisPrint(root);
+    
+    return res;
+};

algorithms/binaryTreePreorderTraversal/binaryTreePreorderTraversal.js

@@ -0,0 +1,34 @@
+var preorderTraversal = function(root) {
+    if (!root) return;
+    const res = [];
+    let cur1 = root;  // 当前开始遍历的节点
+    let cur2;         // 记录当前结点的左子树
+    while(cur1) {
+        cur2 = cur1.left;
+        if(cur2) {
+            // 找到当前左子树的最右侧节点，
+            // 且这个节点应该在指向根结点之前，否则整个节点又回到了根结点。
+            while (cur2.right && cur2.right !== cur1) {
+                cur2 = cur2.right;
+            }
+            // 这个时候如果最右侧这个节点的右指针没有指向根结点，
+            // 创建连接然后往下一个左子树的根结点进行连接操作。
+            if (!cur2.right) {
+                cur2.right = cur1;
+                res.push(cur1.val);
+                cur1 = cur1.left;
+                continue;
+            } else {  // 当左子树的最右侧节点有指向根结点，
+                // 此时说明我们已经回到了根结点并重复了之前的操作，
+                // 同时在回到根结点的时候我们应该已经处理完 左子树的最右侧节点了，把路断开。
+                cur2.right = null;
+            }
+        } else {
+            // 说明当前是叶子节点了，直接访问就好
+            res.push(cur1.val);
+        }
+        // 一直往右边走
+        cur1 = cur1.right;
+    }
+    return res;
+};