3.Tree Traversals (Inorder, Preorder and Postorder)[数据结构]

本文详细介绍了二叉树的三种基本遍历方法:前序遍历、中序遍历和后序遍历,并提供了C和Java语言实现的示例代码,包括递归和非递归两种方式。

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Inorder Traversal:

Algorithm Inorder(tree)
1. Traverse the left subtree, i.e., call Inorder(left-subtree)
2. Visit the root.
3. Traverse the right subtree, i.e., call Inorder(right-subtree)
Uses of Inorder
In case of binary search trees (BST), Inorder traversal gives nodes in non-decreasing order. To get nodes of BST in non-increasing order, a variation of Inorder traversal where Inorder itraversal s reversed, can be used.

Preorder Traversal:

Algorithm Preorder(tree)
1. Visit the root.
2. Traverse the left subtree, i.e., call Preorder(left-subtree)
3. Traverse the right subtree, i.e., call Preorder(right-subtree)
Uses of Preorder
Preorder traversal is used to create a copy of the tree. Preorder traversal is also used to get prefix expression on of an expression tree. Please see http://en.wikipedia.org/wiki/Polish_notation to know why prefix expressions are useful.

Postorder Traversal:

Algorithm Postorder(tree)
1. Traverse the left subtree, i.e., call Postorder(left-subtree)
2. Traverse the right subtree, i.e., call Postorder(right-subtree)
3. Visit the root.
Uses of Postorder
Postorder traversal is used to delete the tree. Please see the question for deletion of tree for details. Postorder traversal is also useful to get the postfix expression of an expression tree. Please see http://en.wikipedia.org/wiki/Reverse_Polish_notation to for the usage of postfix expression.

递归遍历

#include<stdio.h>
#include<stdlib.h>

struct node
{
    int data;
    struct node* left;
    struct node* right;
};

// allocates a new node with given data
struct node* newNode(int data)
{
    struct node* node = (struct node*)malloc(sizeof(struct node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
    return(node);
}

// postorder
void printPostorder(struct node* node)
{
    if(node == NULL)
        return;
    printPostorder(node->left);
    printPostorder(node->right);
    printf("%d ",node->data);
}

// inorder
void printInorder(struct node* node){
    if(node == NULL)
        return;
    printInorder(node->left);
    printf("%d ", node->data);
    printInorder(node->right);
}

//preorder
void printPreorder(struct node* node){
    if(node == NULL)
        return;
    printf("%d ", node->data);
    printPreorder(node->left);
    printPreorder(node->right);
}


int main()
{
    struct node* root = newNode(1);
    root->left = newNode(2);
    root->right           = newNode(3);
     root->left->left     = newNode(4);
     root->left->right   = newNode(5); 

     printf("\nPreorder traversal of binary tree is \n");
     printPreorder(root);

     printf("\nInorder traversal of binary tree is \n");
     printInorder(root);  

     printf("\nPostorder traversal of binary tree is \n");
     printPostorder(root);

     getchar();
     return 0;
}
// Java program for different tree traversals

/* Class containing left and right child of current
   node and key value*/
class Node
{
    int key;
    Node left, right;

    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}

class BinaryTree
{
    // Root of Binary Tree
    Node root;

    BinaryTree()
    {
        root = null;
    }

    /* Given a binary tree, print its nodes according to the
      "bottom-up" postorder traversal. */
    void printPostorder(Node node)
    {
        if (node == null)
            return;

        // first recur on left subtree
        printPostorder(node.left);

        // then recur on right subtree
        printPostorder(node.right);

        // now deal with the node
        System.out.print(node.key + " ");
    }

    /* Given a binary tree, print its nodes in inorder*/
    void printInorder(Node node)
    {
        if (node == null)
            return;

        /* first recur on left child */
        printInorder(node.left);

        /* then print the data of node */
        System.out.print(node.key + " ");

        /* now recur on right child */
        printInorder(node.right);
    }

    /* Given a binary tree, print its nodes in preorder*/
    void printPreorder(Node node)
    {
        if (node == null)
            return;

        /* first print data of node */
        System.out.print(node.key + " ");

        /* then recur on left sutree */
        printPreorder(node.left);

        /* now recur on right subtree */
        printPreorder(node.right);
    }

    // Wrappers over above recursive functions
    void printPostorder()  {     printPostorder(root);  }
    void printInorder()    {     printInorder(root);   }
    void printPreorder()   {     printPreorder(root);  }

    // Driver method
    public static void main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);

        System.out.println("Preorder traversal of binary tree is ");
        tree.printPreorder();

        System.out.println("\nInorder traversal of binary tree is ");
        tree.printInorder();

        System.out.println("\nPostorder traversal of binary tree is ");
        tree.printPostorder();
    }
}

非递归遍历

class Node
{
    int key;
    Node left,right;
    Node(int val)
    {
        key = val;
        left = right = null;
    }
}

class BinaryTree
{
    Node root;
    BinaryTree()
    {
        root = null;
    }

    void printPostorder(Node node)
    {
        if(node == null)
        {
            return;
        }
        printPostorder(node.left);
        printPostorder(node.right);
        System.out.print(node.key+" ");
    }

    void printPostorderNonRecursive(Node root)
    {
        Stack<Node> stack = new Stack<Node>();
        Node currnode = root;
        Node lastvisit = root;
        while(!stack.isEmpty()||currnode!=null){
            while(currnode!=null){
                stack.push(currnode);
                currnode = currnode.left;
            }
            currnode = stack.peek();
            if(currnode.right==null||currnode.right==lastvisit){
                System.out.print(currnode.key+" ");
                stack.pop();
                lastvisit = currnode;
                currnode = null;
            }else{
                currnode = currnode.right;
            }
        }

    }
    /* Given a binary tree, print its nodes in inorder*/
    void printInorder(Node node)
    {
        if (node == null)
            return;

        /* first recur on left child */
        printInorder(node.left);

        /* then print the data of node */
        System.out.print(node.key + " ");

        /* now recur on right child */
        printInorder(node.right);
    }


    void printInorderNonRecursive(Node root)
    {
        Node node  = root;
        Stack<Node> stack = new Stack<Node>();
        while(node!=null||!stack.isEmpty())
        {
            while(node!=null){
                stack.push(node);
                node = node.left;
            }

            if(!stack.isEmpty())
            {
                node = stack.pop();
                System.out.print(node.key+" ");
                node = node.right;
            }
        }
    }


    /* Given a binary tree, print its nodes in preorder*/
    void printPreorder(Node node)
    {
        if (node == null)
            return;

        /* first print data of node */
        System.out.print(node.key + " ");

        /* then recur on left sutree */
        printPreorder(node.left);

        /* now recur on right subtree */
        printPreorder(node.right);
    }

    void printPreorderNonRecursive(Node root)
    {
        Stack<Node> stack = new Stack<Node>();
        stack.push(root);
        while(!stack.isEmpty()){
            Node node = stack.pop();
            if(node.right!=null){
                stack.push(node.right);
            }
            if(node.left!=null){
                stack.push(node.left);
            }
            System.out.print(node.key+" ");
        }
    }




    // Wrappers over above recursive functions
    void printPostorder()  {     printPostorder(root);  }
    void printInorder()    {     printInorder(root);   }
    void printPreorder()   {     printPreorder(root);  }


    public static void main(String[] args){
        BinaryTree tree =  new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);

        System.out.println("Preorder traversal of binary tree is ");
        tree.printPreorder();

        System.out.println("\nInorder traversal of binary tree is ");
        tree.printInorder();

        System.out.println("\nPostorder traversal of binary tree is ");
        tree.printPostorder();
    }

}

http://www.geeksforgeeks.org/tree-traversals-inorder-preorder-and-postorder/

American Heritage 题目描述 Farmer John takes the heritage of his cows very seriously. He is not, however, a truly fine bookkeeper. He keeps his cow genealogies as binary trees and, instead of writing them in graphic form, he records them in the more linear tree in-order" and tree pre-order" notations. Your job is to create the `tree post-order" notation of a cow"s heritage after being given the in-order and pre-order notations. Each cow name is encoded as a unique letter. (You may already know that you can frequently reconstruct a tree from any two of the ordered traversals.) Obviously, the trees will have no more than 26 nodes. Here is a graphical representation of the tree used in the sample input and output: C / \ / \ B G / \ / A D H / \ E F The in-order traversal of this tree prints the left sub-tree, the root, and the right sub-tree. The pre-order traversal of this tree prints the root, the left sub-tree, and the right sub-tree. The post-order traversal of this tree print the left sub-tree, the right sub-tree, and the root. ---------------------------------------------------------------------------------------------------------------------------- 题目大意: 给出一棵二叉树的中序遍历 (inorder) 和前序遍历 (preorder),求它的后序遍历 (postorder)。 输入描述 Line 1: The in-order representation of a tree. Line 2: The pre-o rder representation of that same tree. Only uppercase letter A-Z will appear in the input. You will get at least 1 and at most 26 nodes in the tree. 输出描述 A single line with the post-order representation of the tree. 样例输入 Copy to Clipboard ABEDFCHG CBADEFGH 样例输出 Copy to Clipboard AEFDBHGC c语言,代码不要有注释
最新发布
06-16
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