1 树的序遍历
前序遍历、中序遍历、后序遍历
1.1 遍历方式
都有点抽象，需要结合代码和画图来看
- 递归遍历
- 非递归遍历：都是用栈来解决
- 前序遍历
- 用一个栈，先进右再进左
- 中序遍历
- 用一个栈，先进左，左出，再进右
- 后序遍历
- 用两个栈，一个栈和前序遍历反着来，出的元素进另一个栈，进完之后全打印
- 用一个栈，先解决左边的，再解决右边的。
- 前序遍历
1.2 代码实现
java
public class RecursiveTraversalBT {
public static class Node {
public int val;
public Node left;
public Node right;
public Node(int val){
this.val = val;
}
}
// 递归遍历
public static void pre1(Node head) {
if (head == null) {
return;
}
System.out.print(head.val + " ");
pre1(head.left);
pre1(head.right);
}
public static void in1(Node head) {
if (head == null) {
return;
}
in1(head.left);
System.out.print(head.val + " ");
in1(head.right);
}
public static void pos1(Node head) {
if (head == null) {
return;
}
pos1(head.left);
pos1(head.right);
System.out.print(head.val + " ");
}
// 非递归遍历（栈）
public static void pre2(Node head) {
if (head != null) {
Stack<Node> stack = new Stack<>();
stack.push(head);
while (!stack.isEmpty()) {
head = stack.pop();
System.out.print(head.val + " ");
if (head.right != null) {
stack.push(head.right);
}
if (head.left != null) {
stack.push(head.left);
}
}
}
}
public static void in2(Node head) {
if (head != null) {
Stack<Node> stack = new Stack<>();
while (!stack.isEmpty() || head != null) {
if (head != null) {
stack.push(head);
head = head.left;
}else {
head = stack.pop();
System.out.print(head.val + " ");
head = head.right;
}
}
}
}
public static void pos2(Node head) {
if (head != null) {
Stack<Node> stack1 = new Stack<>();
Stack<Node> stack2 = new Stack<>();
stack1.push(head);
while (!stack1.isEmpty()){
head = stack1.pop();
stack2.push(head);
if (head .left != null) {
stack1.push(head.left);
}
if (head .right != null) {
stack1.push(head.right);
}
}
while (!stack2.isEmpty()) {
System.out.print(stack2.pop().val + " ");
}
}
}
// 后序遍历的第三种写法（只用一个栈）
public static void pos3(Node head) {
if (head != null) {
Stack<Node> stack = new Stack<>();
stack.push(head);
Node help = null;
while (!stack.isEmpty()) {
help = stack.peek();
if (help.left !=null && head != help.left && head != help.right){
stack.push(help.left);
} else if (help.right != null && head != help.right) {
stack.push(help.right);
}else {
System.out.print(stack.pop().val + " ");
head = help;
}
}
}
}
public static void main(String[] args) {
Node head = new Node(1);
head.left = new Node(2);
head.right = new Node(3);
head.left.left = new Node(4);
head.left.right = new Node(5);
head.right.left = new Node(6);
head.right.right = new Node(7);
pos2(head);
System.out.println();
pos3(head);
System.out.println();
}
}2 树的层遍历
2.1 遍历方式
树的最宽层有几个节点？
- 借助Map来查找
- 定义一个队列和一个HashMap，都把头节点放进去，其中Map中存放的是当前节点和其对应的层数
- 每次弹出一个节点，就把这个节点的左右子节点都放进队列和对应的Map
- 如果还没到下一层，那么当前层的节点数就要加一
- 如果到了下一层，就把上一层的节点数和之前层的节点数比较
- 返回节点数最多的层的节点数
- 只用队列
- 定义一个当前层的最左节点，一个下一层的最左节点
- 当弹出的节点等于当前层最左节点时，记录当前层节点数并与之前层的最大节点数比较，哪个大留哪个
- 这个时候就要进入下一次，所以当前层节点数置为0，当前层最左节点置为下一层最左节点
4.2.2 代码实现
java
public class TreeMaxWidth {
public static class Node {
public int val;
public Node left;
public Node right;
public Node(int val) {
this.val = val;
left = null;
right = null;
}
}
public static int maxWidthUseMap(Node head) {
if (head == null) {
return 0;
}
Queue<Node> queue = new LinkedList<>();
queue.add(head);
HashMap<Node, Integer> hashMap = new HashMap<>();
hashMap.put(head, 1);
int curLevelNodes = 0;
int curLevel = 1;
int max = 0;
while (!queue.isEmpty()) {
Node cur = queue.poll();
int curNodeLevel = hashMap.get(cur);
if (cur.left != null) {
queue.add(cur.left);
hashMap.put(cur.left, curNodeLevel + 1);
}
if (cur.right != null) {
queue.add(cur.right);
hashMap.put(cur.right, curNodeLevel + 1);
}
if (curLevel == curNodeLevel) {
curLevelNodes++;
}else {
max = Math.max(max, curLevelNodes);
curLevel++;
curLevelNodes = 1;
}
}
max = Math.max(max, curLevelNodes);
return max;
}
public static int maxWidthNoMap (Node head) {
if (head == null) {
return 0;
}
Queue<Node> queue = new LinkedList<>();
queue.add(head);
Node curEnd = head;
Node nextEnd = null;
int max = 0;
int curLevelNodes = 0;
while (!queue.isEmpty()) {
Node cur = queue.poll();
if (cur.left != null) {
queue.add(cur.left);
nextEnd = cur.left;
}
if (cur.right != null) {
queue.add(cur.right);
nextEnd = cur.right;
}
curLevelNodes++;
if (cur == curEnd) {
max = Math.max(max, curLevelNodes);
curLevelNodes = 0;
curEnd = nextEnd;
}
}
return max;
}
public static void main(String[] args) {
Node head = new Node(1);
head.left = new Node(2);
head.right = new Node(3);
head.left.left = new Node(4);
head.left.right = new Node(5);
head.right.left = new Node(6);
head.right.right = new Node(7);
head.left.right.left = new Node(8);
head.right.left.right = new Node(9);
System.out.println(maxWidthNoMap(head));
}
}