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二叉树的所有遍历完整实现

程序员文章站 2022-05-06 22:53:34
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二叉树的所有遍历完整实现

import java.util.*;

//结点定义
class TreeNode {
    int data;
    TreeNode left, right;

    public TreeNode(int data) {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

//测试
public class Solution {
    public static void main(String args[]) {

        int preOrder[] = new int[]{4, 2, 1, 3, 5, 6};
        int inOrder[] = new int[]{1, 2, 3, 4, 5, 6};
        //调用建树的函数
        Solution s = new Solution();
        TreeNode root = s.buildTree(preOrder, 0, preOrder.length - 1, inOrder, 0, inOrder.length - 1);
        System.out.print("前序递归是:");
       s. preTravel_1(root);
       System.out.println();
        System.out.print("中序递归是:");
        s. inTravel_1(root);
        System.out.println();
        System.out.print("后序递归是:");
        s. postTravel_1(root);
        System.out.println();
        System.out.print("层次遍历是:");
        s.levelTravle(root);
        System.out.println();
        System.out.print("前序非递归是:");
        s. preTravel_2(root);
        System.out.println();
        System.out.print("中序非递归是:");
        s. inTravel_2(root);
        System.out.println();
        System.out.print("后序非递归是:");
      s. postTravel_2(root);
        System.out.println();

    }

    //根据前序和中序建立一颗二叉树
    public TreeNode buildTree(int preOrder[], int preStart, int preEnd, int inOrder[], int inStart, int inEnd) {
        //根节点
        TreeNode root = new TreeNode(preOrder[preStart]);
        //空树,递归出口
        if (preStart == preEnd || inStart == inEnd)
            return root;
        if (preStart > preEnd || inStart > inEnd)
            return null;
        //找到根节点在中序中的位置,以便划分成两部分
        int loc = locateRoot(preOrder[preStart], inOrder);
        root.left = buildTree(preOrder, preStart + 1, loc - inStart + preStart, inOrder, inStart, loc - 1);
        root.right = buildTree(preOrder, loc - inStart + preStart + 1, preEnd, inOrder, loc + 1, inEnd);
        return root;
    }

    //找到结点在中序遍历的位置
    public int locateRoot(int key, int inOrder[]) {
        for (int i = 0; i < inOrder.length; i++)
            if (inOrder[i] == key)
                return i;
        return -1;
    }

    //中序递归


   //前序递归
    public void preTravel_1(TreeNode root) {
        if (root == null)
            return;
        System.out.print(root.data+" ");
        preTravel_1(root.left);
          preTravel_1(root.right);
    }
    //中序递归
    public void inTravel_1(TreeNode root){
        if(root==null)
            return;
        inTravel_1(root.left);
        System.out.print(root.data+" ");
        inTravel_1(root.right);
    }
    //后序递归
    public void postTravel_1(TreeNode root){
        if(root==null)
            return ;
        postTravel_1(root.left);
        postTravel_1(root.right);
        System.out.print(root.data+" ");
    }
    //层次遍历,借助队列
    public void levelTravle(TreeNode root){
        if(root==null)
            return;
        Queue<TreeNode> queue=new LinkedList<>();
        queue.offer(root);
        TreeNode cur=root;
        while(!queue.isEmpty()){
            int n=queue.size();
            for(int i=0;i<n;i++){
               cur = queue.poll();
               System.out.print(cur.data+" ");
              if(cur.left!=null)
                  queue.offer(cur.left);
              if(cur.right!=null)
                  queue.offer(cur.right);
            }

        }

    }
    //前序非递归
    public void preTravel_2(TreeNode root){
      if(root==null)
          return ;
      Stack<TreeNode>stack =new Stack<>();
      TreeNode cur=root;
      stack.push(cur);
      while(!stack.isEmpty()){
          cur=stack.pop();
          System.out.print(cur.data+" ");
          if(cur.right!=null){
              stack.push(cur.right);
          }
          if(cur.left!=null){
              stack.push(cur.left);
          }

      }
    }
    //中序非递归
    public void inTravel_2(TreeNode root){
      if(root==null)
          return ;
      Stack<TreeNode> stack =new Stack<>();
      TreeNode cur=root;
      while(!stack.empty()||cur!=null){
          //先找到最左边的孩子,所有左孩子一次入栈
          while(cur!=null){
              stack.push(cur);
              cur=cur.left;
          }
          //最左边的孩子
          cur=stack.pop();
          System.out.print(cur.data+" ");
          cur=cur.right;
      }



    }
    //后序非递归,这个最难,最好记下来
    public void postTravel_2(TreeNode root){
       if(root==null)
           return;
        Stack<TreeNode> stack=new Stack<>();
        TreeNode cur=root;
        TreeNode pre=null;
        while(!stack.isEmpty()||cur!=null){
            //找到最左边的孩子
            while(cur!=null){
                stack.push(cur);
                cur=cur.left;
            }
            //判断最左边的孩子没有右子树,或者右子树已被访问
            cur=stack.peek();
            if(cur.right==null||cur.right==pre){
                System.out.print(cur.data+" ");
                pre=stack.pop();
                cur=null;              //结束条件

            }else
                cur=cur.right;

        }

    } //end
}






二叉树的所有遍历完整实现
相关标签: 二叉树的遍历