算法设计（迭代后序遍历|S2（使用栈stack）） _后序遍历

本文概述

C
Java
python

我们已经讨论了一个简单的使用两个堆栈进行迭代后遍历在上一篇文章中。在这篇文章中, 讨论了只有一个堆栈的方法。
这个想法是使用左指针向下移动到最左边的节点。向下移动时, 推动根和根的右子堆叠。到达最左边的节点后, 如果没有正确的子节点, 则将其打印出来。如果它有合适的孩子, 则更改根目录, 以便在处理合适的孩子之前进行处理。
以下是详细的算法。

1.1 Create an empty stack 2.1 Do following while root is not NULL a) Push root's right child and then root to stack. b) Set root as root's left child. 2.2 Pop an item from stack and set it as root. a) If the popped item has a right child and the right child is at top of stack, then remove the right child from stack, push the root back and set root as root's right child. b) Else print root's data and set root as NULL. 2.3 Repeat steps 2.1 and 2.2 while stack is not empty.

让我们考虑下面的树

文章图片
以下是使用一个堆栈打印上述树的事后遍历的步骤。

1. Right child of 1 exists. Push 3 to stack. Push 1 to stack. Move to left child. Stack: 3, 12. Right child of 2 exists. Push 5 to stack. Push 2 to stack. Move to left child. Stack: 3, 1, 5, 23. Right child of 4 doesn't exist. ' Push 4 to stack. Move to left child. Stack: 3, 1, 5, 2, 44. Current node is NULL. Pop 4 from stack. Right child of 4 doesn't exist. Print 4. Set current node to NULL. Stack: 3, 1, 5, 25. Current node is NULL. Pop 2 from stack. Since right child of 2 equals stack top element, pop 5 from stack. Now push 2 to stack. Move current node to right child of 2 i.e. 5 Stack: 3, 1, 26. Right child of 5 doesn't exist. Push 5 to stack. Move to left child. Stack: 3, 1, 2, 57. Current node is NULL. Pop 5 from stack. Right child of 5 doesn't exist. Print 5. Set current node to NULL. Stack: 3, 1, 28. Current node is NULL. Pop 2 from stack. Right child of 2 is not equal to stack top element. Print 2. Set current node to NULL. Stack: 3, 19. Current node is NULL. Pop 1 from stack. Since right child of 1 equals stack top element, pop 3 from stack. Now push 1 to stack. Move current node to right child of 1 i.e. 3 Stack: 110. Repeat the same as above steps and Print 6, 7 and 3. Pop 1 and Print 1.

//C program for iterative postorder traversal using one stack #include < stdio.h> #include < stdlib.h> //Maximum stack size #define MAX_SIZE 100//A tree node struct Node { int data; struct Node *left, *right; }; //Stack type struct Stack { int size; int top; struct Node* *array; }; //A utility function to create a new tree node struct Node* newNode( int data) { struct Node* node = ( struct Node*) malloc ( sizeof ( struct Node)); node-> data = https://www.lsbin.com/data; node-> left = node-> right = NULL; return node; }//A utility function to create a stack of given size struct Stack* createStack( int size) { struct Stack* stack = ( struct Stack*) malloc ( sizeof ( struct Stack)); stack-> size = size; stack-> top = -1; stack-> array = ( struct Node**) malloc (stack-> size * sizeof ( struct Node*)); return stack; }//BASIC OPERATIONS OF STACK int isFull( struct Stack* stack) {return stack-> top - 1 == stack-> size; }int isEmpty( struct Stack* stack) {return stack-> top == -1; }void push( struct Stack* stack, struct Node* node) { if (isFull(stack)) return ; stack-> array[++stack-> top] = node; }struct Node* pop( struct Stack* stack) { if (isEmpty(stack)) return NULL; return stack-> array[stack-> top--]; }struct Node* peek( struct Stack* stack) { if (isEmpty(stack)) return NULL; return stack-> array[stack-> top]; }//An iterative function to do postorder traversal of a given binary tree void postOrderIterative( struct Node* root) { //Check for empty tree if (root == NULL) return ; struct Stack* stack = createStack(MAX_SIZE); do { //Move to leftmost node while (root) { //Push root's right child and then root to stack. if (root-> right) push(stack, root-> right); push(stack, root); //Set root as root's left child root = root-> left; }//Pop an item from stack and set it as root root = pop(stack); //If the popped item has a right child and the right child is not //processed yet, then make sure right child is processed before root if (root-> right & & peek(stack) == root-> right) { pop(stack); //remove right child from stack push(stack, root); //push root back to stack root = root-> right; //change root so that the right //child is processed next } else//Else print root's data and set root as NULL { printf ( "%d " , root-> data); root = NULL; } } while (!isEmpty(stack)); }//Driver program to test above functions int main() { //Let us construct the tree shown in above figure struct Node* root = NULL; root = newNode(1); root-> left = newNode(2); root-> right = newNode(3); root-> left-> left = newNode(4); root-> left-> right = newNode(5); root-> right-> left = newNode(6); root-> right-> right = newNode(7); printf ( "Post order traversal of binary tree is :\n" ); printf ( "[" ); postOrderIterative(root); printf ( "]" ); return 0; }

Java

//A java program for iterative postorder traversal using stackimport java.util.ArrayList; import java.util.Stack; //A binary tree node class Node { int data; Node left, right; Node( int item) { data = https://www.lsbin.com/item; left = right; } }class BinaryTree { Node root; ArrayList< Integer> list = new ArrayList< Integer> (); //An iterative function to do postorder traversal //of a given binary tree ArrayList< Integer> postOrderIterative(Node node) { Stack< Node> S = new Stack< Node> (); //Check for empty tree if (node == null ) return list; S.push(node); Node prev = null ; while (!S.isEmpty()) { Node current = S.peek(); /* go down the tree in search of a leaf an if so process it and pop stack otherwise move down */ if (prev == null || prev.left == current || prev.right == current) { if (current.left != null ) S.push(current.left); else if (current.right != null ) S.push(current.right); else { S.pop(); list.add(current.data); }/* go up the tree from left node, if the child is right push it onto stack otherwise process parent and pop stack */ } else if (current.left == prev) { if (current.right != null ) S.push(current.right); else { S.pop(); list.add(current.data); }/* go up the tree from right node and after coming back from right node process parent and pop stack */ } else if (current.right == prev) { S.pop(); list.add(current.data); }prev = current; }return list; }//Driver program to test above functions public static void main(String args[]) { BinaryTree tree = new BinaryTree(); //Let us create trees shown in above diagram 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 ); tree.root.right.left = new Node( 6 ); tree.root.right.right = new Node( 7 ); ArrayList< Integer> mylist = tree.postOrderIterative(tree.root); System.out.println("Post order traversal of binary tree is :" ); System.out.println(mylist); } }//This code has been contributed by Mayank Jaiswal

python

# Python program for iterative postorder traversal # using one stack# Stores the answer ans = []# A Binary tree node class Node:# Constructor to create a new node def __init__( self , data): self .data = https://www.lsbin.com/data self .left = None self .right = Nonedef peek(stack): if len (stack)> 0 : return stack[ - 1 ] return None # A iterative function to do postorder traversal of # a given binary tree def postOrderIterative(root):# Check for empty tree if root is None : return stack = []while ( True ):while (root): # Push root's right child and then root to stack if root.right is not None : stack.append(root.right) stack.append(root)# Set root as root's left child root = root.left# Pop an item from stack and set it as root root = stack.pop()# If the popped item has a right child and the # right child is not processed yet, then make sure # right child is processed before root if (root.right is not None and peek(stack) = = root.right): stack.pop() # Remove right child from stack stack.append(root) # Push root back to stack root = root.right # change root so that the # righ childis processed next# Else print root's data and set root as None else : ans.append(root.data) root = Noneif ( len (stack) < = 0 ): break# Driver pogram to test above function root = Node( 1 ) root.left = Node( 2 ) root.right = Node( 3 ) root.left.left = Node( 4 ) root.left.right = Node( 5 ) root.right.left = Node( 6 ) root.right.right = Node( 7 )print "Post Order traversal of binary tree is" postOrderIterative(root) print ans# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

输出如下：

Post Order traversal of binary tree is [4, 5, 2, 6, 7, 3, 1]

方法2：
向左移动两次, 同时直接推根节点两次。弹出时, 如果发现stack top()与root相同, 则转到root-> right, 否则打印root。

//Simple Java program to print PostOrder Traversal(Iterative) import java.util.Stack; //A binary tree node class Node { int data; Node left, right; Node( int item) { data = https://www.lsbin.com/item; left = right; } }//create a postorder class class PostOrder { Node root; //An iterative function to do postorder traversal //of a given binary tree private void postOrderIterative(Node root) { Stack< Node> stack = new Stack< > (); while ( true ) { while (root != null ) { stack.push(root); stack.push(root); root = root.left; }//Check for empty stack if (stack.empty()) return ; root = stack.pop(); if (!stack.empty() & & stack.peek() == root) root = root.right; else {System.out.print(root.data +" " ); root = null ; } } }//Driver program to test above functions public static void main(String args[]) { PostOrder tree = new PostOrder(); //Let us create trees shown in above diagram 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 ); tree.root.right.left = new Node( 6 ); tree.root.right.right = new Node( 7 ); System.out.println( "Post order traversal of binary tree is :" ); tree.postOrderIterative(tree.root); } }

输出如下：