7 序列和反序列话
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class Solution { /** * This method will be invoked first, you should design your own algorithm * to serialize a binary tree which denote by a root node to a string which * can be easily deserialized by your own "deserialize" method later. */ public String serialize(TreeNode root) { // write your code here if (root == null){ return "{}"; } ArrayList<TreeNode> queue = new ArrayList<>(); queue.add(root); for (int i = 0; i < queue.size(); i++){ TreeNode node = queue.get(i); if (node == null) continue; queue.add(root.left); queue.add(root.right); } while (queue.get(queue.size() - 1) == null){ queue.remove(queue.get(queue.size() - 1)); } StringBuilder sb = new StringBulder(); sb.append("{"); sb.append(queue.get(0)); for (int i = 1; i < queue.size(); i++){ if (queue.get(i) != null){ sb.append(","); sb.append(queue.get(i)); } else{ sb.append(",#"); } } sb.append("}"); return sb.toString(); } /** * This method will be invoked second, the argument data is what exactly * you serialized at method "serialize", that means the data is not given by * system, it's given by your own serialize method. So the format of data is * designed by yourself, and deserialize it here as you serialize it in * "serialize" method. */ public TreeNode deserialize(String data) { // write your code here if ("{}".equals(data)){ return null; } String[] vals = data.substring(1, data.length() - 1).split(","); ArrayList<TreeNode> queue = new ArrayList<>(); TreeNode root = new TreeNode(vals[0]); queue.add(root); boolean isL = true; int index = 0; for (int i = 1; i < vals.length; i++){ if (!"#".equeal(vals[i])){ TreeNode node = new TreeNode(Integer.praiseInt(vals[i])); if (isL){ queue.get(index).left = node; } else{ queue.get(index).right = node; } queue.add(node); } if (!isL){ index++; } isL = !isL; } return root; } }
375 克隆二叉树
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public TreeNode cloneTree(TreeNode root) { // Write your code here if (root == null) return null; TreeNode newR = new TreeNode(root.val); newR.left = cloneTree(root.left); newR.right = cloneTree(root.right); return newR; }
245 子树
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public boolean isSubtree(TreeNode T1, TreeNode T2) { // write your code here if (T2 == null) return true; if (T1 == null) return false; if (Eq(T1, T2)) return true; if (isSubtree(T1.left, T2) || isSubtree(T1.right, T2)) return true; return false; } public boolean Eq(TreeNode t1, TreeNode t2){ if (t1 == null || t2 == null){ return t1 == t2; } if (t1.val != t2.val){ return false; } return Eq(t1.left, t2.left) && Eq(t1.right, t2.right); }
453 展平二叉树
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public void flatten(TreeNode root) { // write your code here LinkedList<TreeNode> pre = new LinkedList<TreeNode>(); pre.add(null); helper(root, pre); } public void helper(TreeNode root, LinkedList<TreeNode> pre) { if (root == null) return; TreeNode right = root.right; if (pre != null){ pre.left = null; pre.right = root; } pre.set(0, root); helper(root.left, pre); helper(right, pre); }
376 二叉树路径和
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public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) { // Write your code here List<List<Integer>> res = new ArrayList<>(); if (root == null) return res; Stack<Integer> stack = new Stack<>(); findPath(root, stack, 0, target, res); return res; } public void findPath(TreeNode root, Stack<Integer> stack, int sum, int target, List<List<Integer>> res){ sum += root.val; stack.push(root.val); if (sum == target && root.left == null && root.right == null){ List<Integer> list = new ArrayList<>(stack); res.add(list); stack.pop(); return; } else{ if (root.left != null){ findPath(root.left, stack, sum,target, res);} if (root.right != null){ findPath(root.right, stack, sum, target, res); } stack.pop(); } }
二叉树路径
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public List<String> binaryTreePaths(TreeNode root) { // Write your code here ArrayList<String> paths = new ArrayList<>(); if (root == null) return paths; List<String> left = binaryTreePaths(root.left); List<String> right = binaryTreePaths(root.right); for (String path : left){ paths.add(root.val + "->" + path); } for (String path : right){ paths.add(root.val + "->" + path); } if (paths.size() == 0){ paths.add("" + root.val); } return paths; }