prólogo
//前序
public static List<Integer> preOrder(TreeNode root){
List<Integer> list = new ArrayList();
Stack<TreeNode> stack = new Stack();
TreeNode cur = root;
while(cur!=null || !stack.isEmpty()){
//一直往左压入栈
while(cur!=null){
list.add(cur.val);
stack.push(cur);
cur = cur.left;
}
cur = stack.pop();
cur = cur.right;
}
return list;
}
Orden medio
//中序
public List<Integer> inorderTraversal(TreeNode root) {
if(root == null){
return new ArrayList();
}
List<Integer> list = new ArrayList();
Stack<TreeNode> stack = new Stack();
TreeNode cur = root;
while(cur != null || !stack.isEmpty()){
while(cur!=null){
stack.push(cur);
cur = cur.left;
}
cur = stack.pop();
list.add(cur.val);
cur = cur.right;
}
return list;
}
Publicar secuencia
//后序遍历,非递归
public static List<Integer> postOrder(TreeNode root){
Stack<TreeNode> stack = new Stack<>();
List<Integer> list = new ArrayList<>();
TreeNode cur = root;
TreeNode p = null;// 用来记录上一节点
while (!stack.isEmpty() || cur != null) {
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
cur = stack.peek();
// 后序遍历的过程中在遍历完左子树跟右子树cur都会回到根结点。
//所以当前不管是从左子树还是右子树回到根结点都不应该再操作了,应该退回上层。
// 如果是从右边再返回根结点,应该回到上层。
// 主要就是判断出来的是不是右子树,是的话就可以把根节点=加入到list了
if (cur.right == null || cur.right == p) {
list.add(cur.val);
stack.pop();
p = cur;
cur = null;
} else {
cur = cur.right;
}
}
return list;
}
nodo
public class TreeNode {
public int val;
public TreeNode left;
public TreeNode right;
public TreeNode(int x) {
val = x;
}
}
Ilustrar el recorrido
En lo anterior, la secuencia transversal es la siguiente:
Recorrido de preorden (centro izquierdo y derecho): 5 4 1 2 6 7 8
Recorrido de orden medio (izquierda, medio y derecha): 1 4 2 5 7 6 8
Recorrido de postorden (izquierda y derecha): 1 2 4 7 8 6 5
modelo
while( 栈非空 || p 非空)
{
if( p 非空)
{
}
else
{
}
}