二叉树递归和非递归遍历
1、递归遍历
// 先序遍历（递归实现）-------------------------------------------------------------
/*
1. Visit the node.
1. Call itself to traverse the node’s left subtree.
3. Call itself to traverse the node’s right subtree.
4. base case: there is no node
*/
private void preOrder(Node localRoot){
if(localRoot != null)
{
visit(localRoot); // System.out.print(localRoot.iData + " ");
preOrder(localRoot.leftChild);
preOrder(localRoot.rightChild);
}
}
// 中序遍历（递归实现）-------------------------------------------------------------
private void inOrder(Node localRoot){
if(localRoot != null)
{
inOrder(localRoot.leftChild);
visit(localRoot); //System.out.print(localRoot.iData + " ");
inOrder(localRoot.rightChild);
}
}
// 后序遍历（递归实现）-------------------------------------------------------------
private void postOrder(Node localRoot){
if(localRoot != null)
{
postOrder(localRoot.leftChild);
postOrder(localRoot.rightChild);
visit(localRoot); //System.out.print(localRoot.iData + " ");
}
}
// -------------------------------------------------------------
2、非递归遍历
/*
总体思想是：因为递归的本质是用栈实现的，所以写出非递归的形式要用到栈。入栈的顺序与访问的顺序相反，如中序遍历访问顺序是left,current,right,所以入栈的顺序相反是right,current,left
steo0:初始化当前节点current
step1:初始化栈
step3:出栈,并访问
Note：中序遍历和后序遍历需要先判断bPushed为true时访问）
step2:入栈
循环结束条件是：栈为空
*/
//算法框架 pushNode是入栈的方式
public void templet(Node root){
//special case
if(root==null)
return;
//inital current node
Node current=root;
//initial stack
Stack s=new Stack();
pushNode(current,s);
//loop: base case is stack is empty
while( !s.empty() ){
//pop
current=s.pop();
//visit or push
}//end while
}//end templet()
// 先序遍历-深度优先搜索（递归实现）-------------------------------------------------------------
//current left right
//step1:pop current node
//step2:visit current node
//step3:push current's right child
//step4:push current's left child
//step5: loop step1-step5
privete void pushPreOrderNode(Node current, Stack s){
if(current==null)
return ;
if(!s.empty()){
leftChild=current.leftChild;
rightChild=curret.rightChild;
if(rightChild!=null) //push rightChild
s.push(rightChild);
if(leftChild!=null) //push leftChild
s.push(leftChild):
visit(current); //current always first visit in preOrder
}//end if
}//end pushPreOrderNode()
public void preOrder(Node root){
//special case
if(root==null)
return ;
//inintial current node
Node current=root;
//inintial stack
Stack s=new Stack();
pushPreOrderNode(current,s);
//base case:stack is empty
while( !s.empty() ){
//pop
current=s.pop();
//push
pushPreOrderNode(current,s);
}//end while
}//end preOrder()
// 中序遍历（非递归实现）-------------------------------------------------------------
/*
left current right
*/
public void inOrder( Node root){
//special case
if(root==null)
return;
Node current=root;
//initial stack
Stack s=new Stack();
pushInOrderNode(current,s);
//pop and visit
while( !s.empty() ){
current=s.pop();
if(current.bPushed)
visit(current);
else
pushInOrderNode(current,s);
}//end while
}//end postOrder()
private void pushInOrderNode(Node current,Stack s){
if(current==null)
return;
if(!s.empty()){
leftChild=current.leftChild;
rightChild=curret.rightChild;
if(rightChild!=null){ //push rightChild
s.push(rightChild);
rightChild.bPushed=false;
}//end if
s.push(current); //push current
if(leftChild!=null){ //push leftChild
s.push(leftChild);
rightChild.bPushed=false;
}//end if
//set flag, ture present current's left and right has both pushed
current.bPushed=true;
}//end if
}//end if
}//end pushInOrderNode()
// -------------------------------------------------------------
// 后序遍历（递归实现）-------------------------------------------------------------
/*
left right current
step1:初始化栈
按照step2初始化栈
step2:入栈
入栈顺序：current,right,left
如果current的中左右节点都入栈了，将current标识为true；
将入栈的左右节点标识初始化为false
step3:出栈
出栈节点设置为current,
如果current标识为ture,访问之；如果为false，做step2步骤
step4:重复step2&step3
*/
public void postOrder( Node root){
//special case
if(root==null)
return;
Node current=root;
//initial stack
Stack s=new Stack();
pushPostOrderNode(current,s);
//pop and visit
while( !s.empty() ){
current=s.pop();
if(current.bPushed)
visit(current);
else
pushPostOrderNode(current,s);
}//end while
}//end postOrder()
// 左右子树尚未入栈，则依次将 根节点,右节点,左节点 入栈
private void pushPostOrderNode(Node current, Stack s){
if(current==null)
return;
if(!s.empty()){
leftChild=current.leftChild;
rightChild=curret.rightChild;
s.push(current); //push current
if(rightChild!=null){ //push rightChild
s.push(rightChild);
rightChild.bPushed=false;
}//end if
if(leftChild!=null){ //push leftChild
s.push(leftChild);
rightChild.bPushed=false;
}//end if
//set flag, ture present current's left and right has both pushed
current.bPushed=true;
}//end if
}//end if
}//end pushPostOrderNode()
3 深度优先搜索和层次遍历（广度优先搜索）
// -------------------------------------------------------------
/*
二叉树的广度深度优先搜索即先序遍历，用栈实现（非递归实现）
*/
// -------------------------------------------------------------
/*
层次搜索，即广度优先搜索
step1:specail case
root==null;
step2:initial current
current=root;
step3:initial queue
step4:remove node in queue, and visit the current node
step5:add node in queue
step6:loop step4 and step5
base case: queue is empty
*/
public void gfs(Noode root){
//step1:specail case
if(root==null)
return ;
//step2:initial current
Node current=root;
//step3:initial queue
Queue q=new Queue();
q.add(current);
//step6:loop step4 and step5
//base case: queue is empty
while( !q.empty() ){
//step4:remove node in queue
current=q.remove();
visit(current);
Node leftChild=current.leftChild;
Node rightChild=curret.rightChild;
//step5:add node in queue
if(leftChild!=null)
q.add(leftChild);
if(rightChild!=null)
q.add(rightChild);
}//end while
}//gfs()
