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\nTo remove the given node，which means whoever is asking the value for this node should get the value of the next node. Plus, the next node of this removing one should be the next next node.
\nIn diagram, with the following node list, we want to remove node 2.
\n1 -> 2 -> 3 -> 4
\nWe update original Node 2 to Node 3 and skip the original Node 3, which gives out the list after removing node 2.
\n1 -> 3 ------> 4
\n/**\n* Definition for singly-linked list.\n* function ListNode(val) {\n* this.val = val;\n* this.next = null;\n* }\n*/\n/**\n* @param {ListNode} node\n* @return {void} Do not return anything, modify node in-place instead.\n*/\nvar deleteNode = function(node) {\n node.val = node.next.val;\n node.next = node.next.next;\n};Click to View the Original Question
\nNote: My solution is not efficient enough based on the performance, but I think it's fairly clean.
\nI use recursion to solve the problem, first to define the basic cases, and recursively loop through both trees from left to right. If it hits one false, the whole result will be false.
/**\n* Definition for a binary tree node.\n* function TreeNode(val) {\n* this.val = val;\n* this.left = this.right = null;\n* }\n*/\n/**\n* @param {TreeNode} p\n* @param {TreeNode} q\n* @return {boolean}\n*/\n \nvar isSameTree = function(p, q) {\n if( p === q ) return true;\n if(!p || !q || p.val !== q.val) return false;\n return isSameTree(p.left, q.left) ? isSameTree(p.right, q.right) : false;\n};Click to View the Original Question
\nHere I have create two indices one called z to index the position for zeroes, another is i to loop through the nums array. They both increment by one when they see a non-zero value, but when it is zero, z stays at position and get ready to swap the position with next non-zero value i meets. By keep doing it, it will fill a consecutive row of zeroes, till the end.
<!--/**\n* @param {number[]} nums\n* @return {void} Do not return anything, modify nums in-place instead.\n*/\nvar moveZeroes = function(nums) {\n var z = 0;\n for(var i = 0; i