算法题解之二叉树与分治法
Different Ways to Add Parentheses
加括号的不同方式
思路:分治地求出一个运算符的左边部分和右边部分的结果,将它们两两组合运算。优化的方式是用一个hash表记录所有字串的中间结果,避免重复计算。
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1 public class Solution { 2 Map<String, List<Integer>> map = new HashMap<String, List<Integer>>(); 3 public List<Integer> diffWaysToCompute(String input) { 4 if (map.containsKey(input)) { 5 return map.get(input); 6 } 7 8 List<Integer> res = new ArrayList<Integer>(); 9 if (!(input.contains("+") || input.contains("-") || input.contains("*"))) { 10 res.add(Integer.parseInt(input)); 11 map.put(input, res); 12 return res; 13 } 14 for (int i = 0; i < input.length(); i++) { 15 char c = input.charAt(i); 16 if (c == '+' || c == '-' || c == '*') { 17 String p1 = input.substring(0, i); 18 String p2 = input.substring(i + 1, input.length()); 19 List<Integer> res1 = diffWaysToCompute(p1); 20 List<Integer> res2 = diffWaysToCompute(p2); 21 22 for (int n1 : res1) { 23 for (int n2 : res2) { 24 if (c == '+') { 25 res.add(n1 + n2); 26 } else if (c == '-') { 27 res.add(n1 - n2); 28 } else if (c == '*') { 29 res.add(n1 * n2); 30 } 31 } 32 } 33 } 34 } 35 map.put(input, res); 36 return res; 37 } 38 }
Invert Binary Tree
反转二叉树
思路1:虽然很简单但是听说homebrew的作者因为没做出这题被谷歌fuck off 了。。所以还是做下吧。。
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1 public class Solution { 2 public TreeNode invertTree(TreeNode root) { 3 if (root == null) { 4 return null; 5 } 6 TreeNode left = root.left; 7 TreeNode right = root.right; 8 root.left = invertTree(right); 9 root.right = invertTree(left); 10 return root; 11 } 12 }
思路2:非递归版。用队列做bfs,把每个节点的左右儿子互换。
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1 public class Solution { 2 public TreeNode invertTree(TreeNode root) { 3 if (root == null) { 4 return null; 5 } 6 Queue<TreeNode> q = new LinkedList<TreeNode>(); 7 q.offer(root); 8 while (!q.isEmpty()) { 9 TreeNode cur = q.poll(); 10 TreeNode left = cur.left; 11 TreeNode right = cur.right; 12 cur.left = right; 13 cur.right = left; 14 if (left != null) { 15 q.offer(left); 16 } 17 if (right != null) { 18 q.offer(right); 19 } 20 } 21 return root; 22 } 23 }
Kth Smallest Element in a BST
二叉树的第k小元素
思路:熟悉二叉树中序遍历的非递归版,这道题就很好做。首先找到最小元素,一定是沿着左子树走到头,然后找次小元素,如果最小节点有右子树,则相当于找右子树的最小元素,如果没有则次小元素为最小元素的父亲。之后的k个元素都这么找。
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1 public class Solution { 2 public int kthSmallest(TreeNode root, int k) { 3 Stack<TreeNode> s= new Stack<TreeNode>(); 4 TreeNode cur = root; 5 while (cur != null) { 6 s.push(cur); 7 cur = cur.left; 8 } 9 10 cur = s.pop(); 11 for (int i = 2; i <= k; i++) { 12 if (cur.right != null) { 13 cur = cur.right; 14 while (cur != null) { 15 s.push(cur); 16 cur = cur.left; 17 } 18 } 19 cur = s.pop(); 20 } 21 return cur.val; 22 } 23 24 }
Lowest Common Ancestor of a Binary Search Tree
二叉查找树的最近公共祖先
思路:与二叉树LCA一样,都是讨论两个节点都在左子树,都在右子树,分别在两边这三种情况。
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1 public class Solution { 2 public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { 3 if (root == null || p == null || q == null) { 4 return null; 5 } 6 if (p.val < root.val && q.val < root.val) { 7 return lowestCommonAncestor(root.left, p, q); 8 } 9 if (p.val > root.val && q.val > root.val) { 10 return lowestCommonAncestor(root.right, p, q); 11 } 12 return root; 13 14 } 15 }