标签：right return offer int null root public
JZ55 二叉树的深度
1 /* 递归 */
2 public class JZ55_1
3 {
4 public static int TreeDepth(TreeNode root)
5 {
6 if (root == null) return 0;
7 return Math.max(TreeDepth(root.left), TreeDepth(root.right)) + 1;
8 }
9 }
10
11 /* 层次遍历 */
12 public class JZ55_2
13 {
14 public static int TreeDepth(TreeNode root)
15 {
16 if (root == null) return 0;
17 Queue<TreeNode> queue = new LinkedList<>();
18 int size = -1, res = 0;
19 queue.add(root);
20 while (!queue.isEmpty())
21 {
22 size = queue.size();
23 while (size != 0)
24 {
25 TreeNode poll = queue.poll();
26 if (poll.left != null)
27 queue.add(poll.left);
28 if (poll.right != null)
29 queue.add(poll.right);
30 size--;
31 }
32 res++;
33 }
34 return res;
35 }
36 }
JZ77 按之字形顺序打印二叉树
1 /* 模拟 */
2 public class JZ77_1
3 {
4 public static ArrayList<ArrayList<Integer>> Print(TreeNode pRoot)
5 {
6 ArrayList<ArrayList<Integer>> res = new ArrayList<>();
7 if (pRoot == null) return res;
8 Queue<TreeNode> queue = new LinkedList<>();
9 boolean revFlag = false;
10 queue.add(pRoot);
11 while (!queue.isEmpty())
12 {
13 ArrayList<Integer> temp = new ArrayList<>();
14 int size = queue.size();
15 while (size != 0)
16 {
17 TreeNode poll = queue.poll();
18 temp.add(poll.val);
19 if (poll.left != null)
20 queue.add(poll.left);
21 if (poll.right != null)
22 queue.add(poll.right);
23 size--;
24 }
25 if (revFlag)
26 Collections.reverse(temp);
27 res.add(temp);
28 revFlag = !revFlag;
29 }
30 return res;
31 }
32 }
33
34 /* 双栈法 */
35 public class JZ77_2
36 {
37 public static ArrayList<ArrayList<Integer>> Print(TreeNode pRoot)
38 {
39 ArrayList<ArrayList<Integer>> res = new ArrayList<>();
40 Stack<TreeNode> stack1 = new Stack<>();
41 Stack<TreeNode> stack2 = new Stack<>();
42 if (pRoot != null)
43 stack1.push(pRoot);
44 while (!stack1.isEmpty() || !stack2.isEmpty())
45 {
46 ArrayList<Integer> temp = new ArrayList<>();
47 if (!stack1.isEmpty())
48 {
49 while (!stack1.isEmpty())
50 {
51 TreeNode popNode = stack1.pop();
52 temp.add(popNode.val);
53 if (popNode.left != null)
54 stack2.push(popNode.left);
55 if (popNode.right != null)
56 stack2.push(popNode.right);
57 }
58 }
59 else
60 {
61 while (!stack2.isEmpty())
62 {
63 TreeNode popNode = stack2.pop();
64 temp.add(popNode.val);
65 if (popNode.right != null)
66 stack1.push(popNode.right);
67 if (popNode.left != null)
68 stack1.push(popNode.left);
69 }
70 }
71 res.add(temp);
72 }
73 return res;
74 }
75 }
JZ54 二叉搜索树的第k个节点
1 /* 递归 */
2 public class JZ54_1
3 {
4 public static int num;
5
6 public static int KthNode(TreeNode proot, int k)
7 {
8 num = k;
9 return inorder(proot);
10 }
11
12 private static int inorder(TreeNode proot)
13 {
14 if (proot != null)
15 {
16 int left = inorder(proot.left);
17 if (left != -1)
18 return left;
19 if (num == 1)
20 return proot.val;
21 num--;
22 return inorder(proot.right);
23 }
24 return -1;
25 }
26 }
27
28 /* 非递归 */
29 public class JZ54_2
30 {
31 public static int KthNode(TreeNode proot, int k)
32 {
33 Stack<TreeNode> stack = new Stack<>();
34 while (!stack.isEmpty() || proot != null)
35 {
36 if (proot != null)
37 {
38 stack.push(proot);
39 proot = proot.left;
40 }
41 else
42 {
43 TreeNode popNode = stack.pop();
44 if (k == 1) return popNode.val;
45 k--;
46 proot = popNode.right;
47 }
48 }
49 return -1;
50 }
51 }
JZ7 重建二叉树
1 /* 递归 */
2 public class JZ7_1
3 {
4 public static TreeNode reConstructBinaryTree(int[] preOrder, int[] vinOrder)
5 {
6 return struct(preOrder, 0, preOrder.length - 1, vinOrder, 0, vinOrder.length - 1);
7 }
8
9 public static TreeNode struct(int[] preOrder, int preS, int preE, int[] vinOrder, int vinS, int vinE)
10 {
11 if (preS > preE || vinS > vinE) return null;
12 TreeNode root = new TreeNode(preOrder[preS]);
13 int pivot = vinS;
14 while (vinOrder[pivot] != preOrder[preS])
15 pivot++;
16 int len = pivot - vinS;
17 root.left = struct(preOrder, preS + 1, preS + len, vinOrder, vinS, pivot - 1);
18 root.right = struct(preOrder, preS + len + 1, preE, vinOrder, pivot + 1, vinE);
19 return root;
20 }
21 }
JZ26 树的子结构⭐
1 /* 递归 */
2 public class JZ26_1
3 {
4 public static boolean HasSubtree(TreeNode root1, TreeNode root2)
5 {
6 if (root2 == null) return false;
7 else return fun(root1, root2);
8 }
9
10 public static boolean fun(TreeNode root1, TreeNode root2)
11 {
12 if (root2 == null) return true;
13 if (root1 == null) return false;
14 boolean one = (root1.val == root2.val) && fun(root1.left, root2.left) && fun(root1.right, root2.right);
15 if (one) return true;
16 boolean two = fun(root1.left, root2) || fun(root1.right, root2);
17 return two;
18 }
19 }
20
21 /* 递归版本2 */
22 public class JZ26_2
23 {
24 public static boolean HasSubtree(TreeNode root1, TreeNode root2)
25 {
26 boolean res = false;
27 if (root1 != null && root2 != null)
28 {
29 if (root1.val == root2.val)
30 res = judge(root1, root2);
31 if (!res)
32 res = HasSubtree(root1.left, root2);
33 if (!res)
34 res = HasSubtree(root1.right, root2);
35 }
36 return res;
37 }
38
39 public static boolean judge(TreeNode root1, TreeNode root2)
40 {
41 if (root2 == null) return true;
42 if (root1 == null) return false;
43 if (root1.val != root2.val) return false;
44 return judge(root1.left, root2.left) && judge(root1.right, root2.right);
45 }
46 }
JZ27 二叉树的镜像
1 /* 递归 */
2 public class JZ7_1
3 {
4 public static TreeNode reConstructBinaryTree(int[] preOrder, int[] vinOrder)
5 {
6 return struct(preOrder, 0, preOrder.length - 1, vinOrder, 0, vinOrder.length - 1);
7 }
8
9 public static TreeNode struct(int[] preOrder, int preS, int preE, int[] vinOrder, int vinS, int vinE)
10 {
11 if (preS > preE || vinS > vinE) return null;
12 TreeNode root = new TreeNode(preOrder[preS]);
13 int pivot = vinS;
14 while (vinOrder[pivot] != preOrder[preS])
15 pivot++;
16 int len = pivot - vinS;
17 root.left = struct(preOrder, preS + 1, preS + len, vinOrder, vinS, pivot - 1);
18 root.right = struct(preOrder, preS + len + 1, preE, vinOrder, pivot + 1, vinE);
19 return root;
20 }
21 }
JZ32 从上往下打印二叉树
1 /* 层次遍历 */
2 public class JZ32_1
3 {
4 public static ArrayList<Integer> PrintFromTopToBottom(TreeNode root)
5 {
6 ArrayList<Integer> res = new ArrayList<>();
7 Queue<TreeNode> queue = new LinkedList<>();
8 if (root == null) return res;
9 queue.add(root);
10 while (!queue.isEmpty())
11 {
12 TreeNode poll = queue.poll();
13 res.add(poll.val);
14 if (poll.left != null)
15 queue.add(poll.left);
16 if (poll.right != null)
17 queue.add(poll.right);
18 }
19 return res;
20 }
21 }
22
23 /* 递归 */
24 public class JZ32_2
25 {
26 public static ArrayList<Integer> PrintFromTopToBottom(TreeNode root)
27 {
28 ArrayList<ArrayList<Integer>> temp = new ArrayList<>();
29 ArrayList<Integer> res = new ArrayList<>();
30 levelTravel(root, temp, 1);
31 for (ArrayList<Integer> list : temp)
32 res.addAll(list);
33 return res;
34 }
35
36 public static void levelTravel(TreeNode root, ArrayList<ArrayList<Integer>> temp, int depth)
37 {
38 if (root == null) return;
39 if (temp.size() < depth)
40 {
41 ArrayList<Integer> addList = new ArrayList<>();
42 addList.add(root.val);
43 temp.add(addList);
44 }
45 else
46 {
47 temp.get(depth - 1).add(root.val);
48 }
49 levelTravel(root.left, temp, depth + 1);
50 levelTravel(root.right, temp, depth + 1);
51 }
52 }
JZ33 二叉搜索树的后序遍历序列
1 /* 递归 */
2 public class JZ33_1
3 {
4 public static boolean VerifySquenceOfBST(int[] sequence)
5 {
6 if (sequence.length == 0) return false;
7 return judge(sequence, 0, sequence.length - 1);
8 }
9
10 public static boolean judge(int[] sequence, int left, int right)
11 {
12 if (left >= right) return true;
13 int idx = left;
14 while (sequence[idx] < sequence[right] && idx < right)
15 idx++;
16 for (int i = idx; i < right; i++)
17 {
18 if (sequence[i] < sequence[right])
19 return false;
20 }
21 return judge(sequence, left, idx - 1) && judge(sequence, idx, right - 1);
22 }
23 }
24
25 /* ⭐若将中序序列当作入栈序列，那么后序序列应该是合法的出栈序列⭐ */
26 public class JZ33_2
27 {
28 public static boolean VerifySquenceOfBST(int[] sequence)
29 {
30 if (sequence.length == 0) return false;
31 int j = 0;
32 ArrayList<Integer> inOrder = new ArrayList<>();
33 Stack<Integer> stack = new Stack<>();
34 for (int num : sequence) inOrder.add(num);
35 Collections.sort(inOrder);
36 for (int num : inOrder)
37 {
38 stack.add(num);
39 while (!stack.isEmpty() && stack.peek() == sequence[j])
40 {
41 stack.pop();
42 j++;
43 }
44 }
45 return j == sequence.length;
46 }
47 }
JZ82 二叉树中和为某一值的路径(一)
1 /* 递归 */
2 public class JZ82_1
3 {
4 public static boolean hasPathSum(TreeNode root, int sum)
5 {
6 if (root == null)
7 return false;
8 if (root.left == null && root.right == null && sum - root.val == 0)
9 return true;
10 return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
11 }
12 }
13
14 /* 层次 */
15 public class JZ82_2
16 {
17 public static boolean hasPathSum(TreeNode root, int sum)
18 {
19 if (root == null) return false;
20 Queue<TreeNode> queue = new LinkedList<>();
21 queue.add(root);
22 while (!queue.isEmpty())
23 {
24 TreeNode pollNode = queue.poll();
25 if (pollNode.left != null)
26 {
27 pollNode.left.val += pollNode.val;
28 queue.add(pollNode.left);
29
30 }
31 if (pollNode.right != null)
32 {
33 pollNode.right.val += pollNode.val;
34 queue.add(pollNode.right);
35 }
36 if (pollNode.val == sum && pollNode.left == null && pollNode.right == null)
37 return true;
38 }
39 return false;
40 }
41 }
JZ34 二叉树中和为某一值的路径(二)
1 /* DFS */
2 public class JZ34_1
3 {
4 public static ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
5
6 public static ArrayList<ArrayList<Integer>> FindPath(TreeNode root, int target)
7 {
8 if (root == null) return res;
9 dfs(root, target, new ArrayList<Integer>());
10 return res;
11 }
12
13 public static void dfs(TreeNode root, int target, ArrayList<Integer> path)
14 {
15 path.add(root.val);
16 target -= root.val;
17 if (root.left == null && root.right == null && target == 0)
18 res.add(new ArrayList<>(path));
19 if (root.left != null)
20 dfs(root.left, target, path);
21 if (root.right != null)
22 dfs(root.right, target, path);
23 path.remove(path.size() - 1);
24 }
25 }
标签：right,
return,
offer,
int,
null,
root,
public
From： https://www.cnblogs.com/VividBinGo/p/17721156.html