标签:poll oid comment his 左右子树 pre linked 出队 ams
我用下图的树为例,做树的遍历:

树节点的定义:
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;
    public TreeNode(int val) {
        this.val = val;
    }
    public TreeNode(int val, TreeNode left, TreeNode right) {
        super();
        this.val = val;
        this.left = left;
        this.right = right;
    }}树的结构的代码实现:public static void main(String[] args) {
       TreeNode e = new TreeNode(1);
       TreeNode g = new TreeNode(2);
       TreeNode h = new TreeNode(3);
       TreeNode i = new TreeNode(4);
       TreeNode d = new TreeNode(5,null,g);
       TreeNode f = new TreeNode(6,h,i); 
       TreeNode b = new TreeNode(7,d,e);
       TreeNode c = new TreeNode(8,f,null);
       TreeNode root = new TreeNode(9,b,c);}
public static void printTree(TreeNode t){
       if(t!=null){
           printTree(t.left);
           System.out.print(t.val+" ");
           printTree(t.right);
       }
   }
输出结果:
5 2 7 1 9 3 6 4 8
public static void printTree(TreeNode t){
        if(t!=null){
            printTree(t.left);
            printTree(t.right);
            System.out.print(t.val+" ");
        }
    }
输出结果:
2 5 1 7 3 4 6 8 9
public static void printTree(TreeNode t){
        if(t!=null){
            System.out.print(t.val+" ");
            printTree(t.left);
            printTree(t.right);
        }
    }
输出结果:
9 7 5 2 1 8 6 3 4
有没有觉得树的先序,中序,后序遍历都非常简单,递归三行代码就搞定了。好吧,下边厉害的要来了
层序遍历:所有深度为D的节点要在深度为D+1的节点之前进行处理,层序遍历与其他类型的遍历不同的地方在于它不是递归地执行的,它用到队列,而不使用递归所默示的栈。
算法思想:
package Tree;import java.util.ArrayList;import java.util.LinkedList;import java.util.List;import java.util.Queue;/* * 层序遍历 *  */public class TreePrinter1 {
     public static int[][] printTree(TreeNode root) {
         List< List<Integer> > list = new ArrayList< List<Integer> >();
         list.add(new ArrayList<Integer>());
         Queue<TreeNode> queue = new LinkedList<TreeNode>();
            queue.add(root);
            TreeNode lastNode = root;    // 当前行最右节点
            TreeNode nlastNode = root;    // 下一行最右节点
            TreeNode tmpNode = null; 
            int hight = 0;                // 树的高度
            while(!queue.isEmpty()){
                tmpNode = queue.poll();
                if(tmpNode!=null){
                    list.get(hight).add(tmpNode.val);
                }
                if(tmpNode.left!=null){
                    queue.add(tmpNode.left);
                    nlastNode = tmpNode.left;
                }
                if(tmpNode.right!=null){
                    queue.add(tmpNode.right);
                    nlastNode = tmpNode.right;
                }
                if(tmpNode == lastNode){
                    lastNode = nlastNode;
                    hight++;
                    list.add(new ArrayList<Integer>());
                }
            }
            int[][] data = new int[list.size()][];
            for(int i=0;i<list.size();i++){
                for(int j=0;j<list.get(i).size();j++){
                    data[i][j] = list.get(i).get(j);
                }
            }
        return data;
     }
     public static void main(String[] args) {
           TreeNode e = new TreeNode(1);
           TreeNode g = new TreeNode(2);
           TreeNode h = new TreeNode(3);
           TreeNode i = new TreeNode(4);
           TreeNode d = new TreeNode(5,null,g);
           TreeNode f = new TreeNode(6,h,i);
           TreeNode b = new TreeNode(7,d,e);
           TreeNode c = new TreeNode(8,f,null);
           TreeNode root = new TreeNode(9,b,c);
           int[][] data  =TreePrinter.printTree(root);
           for(int s=0;s<data.length;s++){
               for(int j=0;j<data[s].length;j++){
                   System.out.print(data[s][j]+" ");
               }
               System.out.println();
           }
    }}
输出结果:
9
7 8
5 1 6
2 3 4
标签:poll oid comment his 左右子树 pre linked 出队 ams
原文地址:https://www.cnblogs.com/bianchengrufeng/p/11152890.html