问题
Can someone please explain to me how recursion works in in Order traversal. here's my inOrder() method.
public void inOrder(BinaryNode p){
if(p.left!=null){
inOrder(p.left);
}
visit(p);
if(p.right!=null){
inOrder(p.right);
}
}
public void visit(BinaryNode p){
System.out.println(p.element);
}
BinaryTree t=new BinaryTree();
t.insert(5);
t.insert(t.root,4);
t.insert(t.root,6);
t.insert(t.root,60);
t.insert(t.root,25);
t.insert(t.root,10);
t.inOrder(t.root);
The method inOrder() prints the elements correctly,but I don't understand how it works.
When I call t.inOrder(t.root); since root has value 5 it would be similar to inOrder(5);
and that has a left node so if(p.left!=null){
inOrder(p.left);
}
would get executed.There the recursion call would be inOrder(4);
Since 4's left points to null, then visit(4) is the line that executed printing the value 4.
But then after that how does 5 get printed.Although at first when the method was called by t.inOrder(t.root); the local variable p was assigned with BinaryNode of value 5, now p is 4. Then after printing out 4, the next line that can get executed is
if(p.right!=null){
inOrder(p.right);
}
But since p.right now refers to right in BinaryNode with element 4 and 4's right is null, this also won't get executed.
Then how does the recursion is maintained?
How does it print out 5 and the rest of the nodes?
回答1:
This is hard to say.. It depend on your implementation..
I added implementation with in order traverse.. Hope that helps
class BinaryTreeSearch{
public enum State{
Visited, Unvisited,Visiting;
}
//this is the Node used in the tree
static class Node{
private int data;
private Node left;
private Node right;
public Node(int data){
this.data = data;
left = null;
right = null;
}
public void setLeft(Node left){
this.left = left;
}
public void setRight(Node right){
this.right = right;
}
public Node getLeft(){
return this.left;
}
public Node getRight(){
return this.right;
}
public int getData(){
return this.data;
}
public boolean equals(Node n){
if(this.data ==(int) n.getData()) return true;
else
return false;
}
}
public static void main(String[] args){
BinaryTreeSearch bts = new BinaryTreeSearch();
bts.run();
}
//execute the test case
public void run(){
Node root = new Node(10);
insert(root,new Node(20));
insert(root,new Node(5));
insert(root,new Node(4));
insert(root,new Node(5));
insert(root,new Node(15));
inOrderTraverse(root);
System.out.println("\n" + binarySearch(root,new Node(10)));
}
// insert a node to the binary search tree
public void insert(Node root, Node n){
if(root == null|| n == null) return;
if(root.getData() > n.getData()){
if(root.getLeft() == null){
root.setLeft(n);
System.out.println("Added node to left of "+root.getData()+" of value "+n.getData());
}else{
insert(root.getLeft(),n);
}
}else if(root.getData() < n.getData()){
if(root.getRight() == null){
root.setRight(n);
System.out.println("Added node to Right of "+root.getData()+" of value "+n.getData());
}else{
insert(root.getRight(),n);
}
}
}
//in-order Traversal
public void inOrderTraverse(Node root){
if(root != null){
inOrderTraverse(root.getLeft());
System.out.print(" "+root.getData());
inOrderTraverse(root.getRight());
}
}
//binary search
public boolean binarySearch(Node root,Node n){
if(root == null || n == null) {
return false;
}
System.out.println(" Testing out "+root.getData()+" for value "+n.getData());
if(root.getData() > n.getData()){
return binarySearch(root.getLeft(),n);
}else if(root.getData() < n.getData()){
return binarySearch(root.getRight(),n);
}
return true;
}
}
回答2:
I've explained the best i could without images. do print(i) means printing i and do inorder(i) means inorder(i) is expanded to inorder(left of i) > print (i) > inorder(right of i)
inorder(5) called
todo: inorder(4) > print 5 > inorder(6)
do inorder(4)
todo: inorder(left of 4)=nothing > print(4) > inorder(right of 4)=nothing > print(5)
inorder 6
do print 4
do print 5
do inorder 6
todo: inorder(left of 6)=nothing > print 6 > inorder(60)
do print 6
do inorder 60
todo: inorder(25) > print 60 > inorder(right of 60)=nothing
do inorder 25
todo: inorder(10) > print 25 > inorder(right of 25)=nothing > print 60
do inorder 10
todo: inorder(left of 10)=nothing > print 10 >inorder(right of 10)=nothing >print 25>print 60
do print 10
do print 25
do print 60
So if you see the order of printing its 4 5 6 10 25 60
来源：https://stackoverflow.com/questions/23852798/in-order-recursion-in-binary-trees