总结一下，二叉树的各种遍历的递归和非递归实现，以及层次遍历（每一层输出一行）
实现代码如下：
package stone.offer.scu;
import java.awt.HeadlessException;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
//二叉树的遍历
public class TraversalBiTree {
public static void main(String[] args) {
//以数组形式生成一颗完全二叉树
TreeNode[] treeNodes = new TreeNode[10];
for (int i = 0; i < 10; i++) {
treeNodes[i] = new TreeNode(i);
}
for (int i = 0; i < 10; i++) {
if (i*2+1 < 10) {
treeNodes[i].left = treeNodes[i*2+1];
}
if (i*2+2 < 10) {
treeNodes[i].right = treeNodes[i*2+2];
}
}
// PreOrder(treeNodes[0]);
// postOrderRe(treeNodes[0]);
// postOrder(treeNodes[0]);
LayerTravel(treeNodes[0]);
}
//前序遍历--递归
public static void PreOrderRe(TreeNode treeNode) {
if (treeNode == null) {
return;
}else {
System.out.println(treeNode.val);
PreOrderRe(treeNode.left);
PreOrderRe(treeNode.right);
}
}
//前序遍历--非递归借助栈
public static void PreOrder(TreeNode treeNode) {
Stack<TreeNode> stack = new Stack<>();
System.out.println("前序非递归：");
while (treeNode != null || !stack.isEmpty()) {
while (treeNode != null) {
System.out.print(treeNode.val + " ");
stack.push(treeNode);
treeNode = treeNode.left;
}
if (!stack.isEmpty()) {
treeNode = stack.pop();
treeNode = treeNode.right;
}
}
}
//中序遍历-递归
public static void midOrderRe(TreeNode treeNode) {
if (treeNode == null) {
return;
}else {
midOrderRe(treeNode.left);
System.out.println(treeNode.val);
midOrderRe(treeNode.right);
}
}
//中序遍历-非递归
public static void midOrder(TreeNode treeNode) {
Stack<TreeNode> stack = new Stack<>();
while (treeNode != null || !stack.isEmpty()) {
while (treeNode != null) {
stack.push(treeNode);
treeNode = treeNode.left;
}
if (!stack.isEmpty()) {
treeNode = stack.pop();
System.out.println(treeNode.val);
treeNode = treeNode.right;
}
}
}
//后序遍历--递归
public static void postOrderRe(TreeNode treeNode) {
if (treeNode == null) {
return;
}else {
postOrderRe(treeNode.left);
postOrderRe(treeNode.right);
System.out.print(treeNode.val + " ");
}
}
//后序遍历-非递归
// 后序遍历递归定义：先左子树，后右子树，再根节点。
// 后序遍历的难点在于：需要判断上次访问的节点是位于左子树，还是右子树。
public static /*ArrayList<Integer>*/ void postOrder(TreeNode treeNode) {
// ArrayList<Integer> res = new ArrayList<>();
if (treeNode == null) {
// return res;
return;
}
Stack<TreeNode> stack = new Stack<>();
TreeNode cur = treeNode; //用来记录当前访问结点
TreeNode pre = null; // 记录上次访问结点
//先把左枝全入栈
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
while (!stack.isEmpty()) {
cur = stack.pop();
//如果当前访问结点的右孩子为空 或 与上次访问结点相同（说明顺序为：右孩子->根）
if (cur.right == null || cur.right == pre) {
// res.add(cur.val);
System.out.print(cur.val + " "); //访问
pre = cur; //更新上次访问结点
}else {
//如果为根结点，再次入栈
stack.push(cur);
cur = cur.right; //再将其右孩子入栈
while (cur != null) {
stack.push(cur);
cur = cur.left; //继续放入右孩子的左枝
}
}
}
// return res;
}
//层次遍历 队列
public static void LayerTravel(TreeNode treeNode) {
if (treeNode == null) {
return;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.add(treeNode);
while (!queue.isEmpty()) {
TreeNode cur = queue.poll();
System.out.print(cur.val + " ");
if (cur.left != null) {
queue.add(cur.left);
}
if (cur.right != null) {
queue.add(cur.right);
}
}
}
//层次遍历-- 队列 单行打印
public static ArrayList<ArrayList<Integer>> layerTravel(TreeNode treeNode) {
ArrayList<Integer> layer = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList<>();
ArrayList<ArrayList<Integer>> res = new ArrayList<>();
if (treeNode == null) {
return res;
}
queue.add(treeNode);
int start = 0,end = 1;
while (!queue.isEmpty()) {
TreeNode cur = queue.poll();
layer.add(cur.val);
start++;
if (cur.left != null) {
queue.add(cur.left);
}
if (cur.right != null) {
queue.add(cur.right);
}
if (start == end) {
end = queue.size(); //下一层数量
start = 0; //start清零
res.add(layer);
layer = new ArrayList<Integer>();
}
}
return res;
}
}