Leetcode刷題--Week 3

鏈表插入排序

題目

Sort a linked list using insertion sort.

解題思路

1. 一個指針指從原始列表頭部開始逐步往後移指向插入元素；
2. 一個指針從頭遍歷已排序鏈表，找到插入節點，將元素插入：

代碼

 if (head == null || head.next == null){
			return head;
		}
		ListNode first = new ListNode(0);
		first.next = head;
		ListNode cur = head.next;
		head.next = null;
		while (cur != null){
			ListNode p = first;
			while(p!=null && p.next != null){
				if (cur.val < p.next.val ){
					break;
				}
				p = p.next;
			}
			ListNode temp = cur.next;
			cur.next = p.next;
			p.next = cur;
			cur = temp;
		}
		return first.next;

二叉樹後序遍歷

題目

Given a binary tree, return the postorder traversal of its nodes’ values.
For example:
Given binary tree{1,#,2,3}, return[3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?

解法一，遞歸遍歷

遞歸遍歷比較簡單，直接上代碼

代碼

    public ArrayList<Integer> postorderTraversal(TreeNode root) {
		ArrayList<Integer> orderList = new ArrayList<>();
		traversal(orderList,root);
		return orderList;
	}
	
    public void traversal(ArrayList<Integer> list, TreeNode node){
		if (node == null ){
			return;
		}
		traversal(list, node.left);
		traversal(list, node.right);
		list.add(node.val);
	}

解法二，非遞歸遍歷

1. 將根節點如棧
2. 循環訪問棧頂節點，若棧頂節點是葉子節點或是上次出棧節點的父節點，將其值放入列表中，並將該元素出棧；
3. 若右節點不爲空則將右節點入棧，左節點不爲空也將左節點入棧

代碼

    public ArrayList<Integer> postorderTraversalItera(TreeNode root) {
		ArrayList<Integer> orderList = new ArrayList<>();
		Stack<TreeNode> stack = new Stack<>();
		if (root != null){
			stack.push(root);
		}
		TreeNode head = root;
		while (!stack.empty()){
			TreeNode p = stack.peek();
			if ( (p.left == null && p.right == null) || p.right == head || p.left == head ){
				orderList.add(p.val);
				head = p;
				stack.pop();
			}else {

				if (p.right != null){
					stack.push(p.right);
				}

				if (p.left != null){
					stack.push(p.left);
				}

			}
		}
		return orderList;
	}