树的两种遍历

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
1. 前序遍历
  - 用一个栈，先进右再进左
2. 中序遍历
  - 用一个栈，先进左，左出，再进右
3. 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。

1.2 代码实现

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
1. 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
2. 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
3. 如果还没到下一层，那么当前层的节点数就要加一
4. 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
5. 返回节点数最多的层的节点数
只用队列
1. 定义一个当前层的最左节点，一个下一层的最左节点
2. 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
3. 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

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));}
}