1.二叉树的层序遍历
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new LinkedList<>();
if (root==null) return res;
Queue<TreeNode> tree = new LinkedList<>();
tree.add(root);
while (!tree.isEmpty()){
int size = tree.size();
List<Integer> ans = new LinkedList<>();
for (int i = 0;i<size;i++){
TreeNode temp = tree.remove();
ans.add(temp.val);
if (temp.left!=null) tree.add(temp.left);
if (temp.right!=null) tree.add(temp.right);
}
res.add(ans);
}
return res;
}
}
2.二叉树的最大深度
class Solution {
private int maxdepth = 0;
public int maxDepth(TreeNode root) {
if (root==null) return 0;
int depth = Depth(root,1);
return depth;
}
private int Depth(TreeNode root, int i) {
maxdepth = Math.max(i, maxdepth);
if (root.left!=null) Depth(root.left, i+1);
if (root.right!=null) Depth(root.right, i+1);
return maxdepth;
}
}
class Solution {
public int maxDepth(TreeNode root) {
if (root==null) return 0;
int depth_left = maxDepth(root.left);
int depth_right = maxDepth(root.right);
return Math.max(depth_left, depth_right)+1;
}
}
3.对称二叉树
class Solution {
public boolean isSymmetric(TreeNode root) {
if (root==null) return true;
return judge(root.left,root.right);
}
private boolean judge(TreeNode left, TreeNode right) {
if (left==null&&right==null) return true;
if (left==null||right==null) return false;
boolean res = (left.val==right.val)&&(judge(left.right, right.left)&&
judge(left.left, right.right));
return res;
}
}
4.路径总和
class Solution {
private int judge = 0;
public boolean hasPathSum(TreeNode root, int sum) {
if(root==null) return false;
return DFS(root,sum,root.val)==1?true:false;
}
private int DFS(TreeNode root, int sum, int temp) {
if (root.left==null&&root.right==null&&temp==sum) judge = 1;
if (root.left!=null) DFS(root.left, sum, temp+root.left.val);
if (root.right!=null) DFS(root.right, sum, temp+root.right.val);
return judge;
}
}
5.从中序与后序遍历序列构造二叉树
class Solution {
HashMap<Integer,Integer> inorderMap = new HashMap<>();
int[] post;
public TreeNode buildTree(int[] inorder, int[] postorder) {
post = postorder;
for (int i =0 ;i<inorder.length;i++){
inorderMap.put(inorder[i],i);
}
TreeNode root = create(0,inorder.length-1,0,postorder.length-1);
return root;
}
private TreeNode create(int inorderbegin, int inorderend, int postorderbegin, int postorderend) {
if (inorderbegin>inorderend||postorderbegin>postorderend) return null;
int rootNum = inorderMap.get(post[postorderend]);
TreeNode node = new TreeNode(post[postorderend]);
node.left = create(inorderbegin,rootNum-1,postorderbegin,postorderbegin+rootNum-1-inorderbegin);
node.right = create(rootNum+1, inorderend, postorderbegin+rootNum-inorderbegin, postorderend-1);
return node;
}
}
6.从前序与中序遍历序列构造二叉树&&重建二叉树
class Solution {
HashMap<Integer,Integer> inorderMap = new HashMap<>();
int[] pre;
public TreeNode buildTree(int[] preorder, int[] inorder) {
pre = preorder;
for (int i = 0;i<inorder.length;i++){
inorderMap.put(inorder[i],i);
}
TreeNode root = create(0,inorder.length-1,0, preorder.length-1);
return root;
}
private TreeNode create(int inorderbegin, int inorderend, int preorderbegin, int preorderend) {
if (inorderbegin>inorderend||preorderbegin>preorderend) return null;
int rootNum = inorderMap.get(pre[preorderbegin]);
TreeNode node = new TreeNode(pre[preorderbegin]);
node.left = create(inorderbegin,rootNum-1,preorderbegin+1,preorderbegin+rootNum-inorderbegin);
node.right = create(rootNum+1, inorderend, preorderbegin+rootNum-inorderbegin+1, preorderend);
return node;
}
}
7.从前序与后序遍历序列构造二叉树
class Solution {
HashMap<Integer,Integer> postorderMap = new HashMap<>();
int[] preorder ;
public TreeNode constructFromPrePost(int[] pre, int[] post) {
preorder = pre;
for(int i = 0;i<pre.length;i++){
postorderMap.put(post[i],i);
}
TreeNode root = create(0,pre.length-1,0,post.length-1);
return root;
}
private TreeNode create(int prebegin, int preend, int postbegin, int postend) {
if(prebegin == preend) return new TreeNode(preorder[prebegin]);
if (prebegin>preend||postbegin>postend)return null;
TreeNode node = new TreeNode(preorder[prebegin]);
int temp = postorderMap.get(preorder[prebegin+1]);
int leftnum = temp-postbegin+1;
node.left =create(prebegin+1,prebegin+leftnum,
postbegin,temp);
node.right = create(prebegin+leftnum+1, preend, temp+1, postend-1);
return node;
}
}
8.填充每个节点的下一个右侧节点指针
public Node connect(Node root) {
if (root==null) return null;
Queue<Node> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()){
int size = queue.size();
for (int i = 0;i<size;i++){
Node temp = queue.poll();
if (i<size-1) temp.next = queue.peek();
if (temp.left!=null) queue.add(temp.left);
if (temp.right!=null) queue.add(temp.right);
}
}
return root;
}
}
9.填充每个节点的下一个右侧节点指针 II
public Node connect(Node root) {
if (root==null) return null;
Queue<Node> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()){
int size = queue.size();
for (int i = 0;i<size;i++){
Node temp = queue.poll();
if (i<size-1) temp.next = queue.peek();
if (temp.left!=null) queue.add(temp.left);
if (temp.right!=null) queue.add(temp.right);
}
}
return root;
}
}
10.二叉树的最近公共祖先
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root==null||root==p||root==q) return root;
TreeNode judgeleft = lowestCommonAncestor(root.left, p, q);
TreeNode judgeright = lowestCommonAncestor(root.right, p, q);
if (judgeleft!=null&&judgeright!=null) return root;
else if (judgeleft==null) return judgeright;
else if (judgeright==null) return judgeleft;
return null;
}
}
11.序列化和反序列化二叉搜索树
class Solution {
public class Codec {
public String serialize(TreeNode root) {
StringBuilder string = new StringBuilder();
serialize(string, root);
return string.toString();
}
private void serialize(StringBuilder string, TreeNode root) {
if (root == null) {
string.append("#").append(",");
return;
}
string.append(root.val).append(",");
serialize(string, root.left);
serialize(string, root.right);
}
public TreeNode deserialize(String data) {
LinkedList<String> nodes = new LinkedList<>();
for (String temp : data.split(",")) {
nodes.add(temp);
}
return deserialize(nodes);
}
private TreeNode deserialize(LinkedList<String> nodes) {
if (nodes.isEmpty()) return null;
String node = nodes.removeFirst();
if (node.equals("#")) return null;
TreeNode root = new TreeNode(Integer.parseInt(node));
root.left = deserialize(nodes);
root.right = deserialize(nodes);
return root;
}
}
}