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TraverseOrder.java
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219 lines (209 loc) · 6.48 KB
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package basic.tree;
import java.util.*;
/**
* @author JunjunYang
* @date 2020/1/1 12:10
*/
public class TraverseOrder {
/**
* 递归前序遍历
*
* @param root
* @return
*/
public static List<Integer> preOrderRecursion(TreeNode root) {
List<Integer> list = new ArrayList<>();
preOrderRecursion(root, list);
return list;
}
private static void preOrderRecursion(TreeNode root, List<Integer> ans) {
if (root != null) {
ans.add(root.val);
preOrderRecursion(root.left,ans);
preOrderRecursion(root.right,ans);
}
}
/**
* 非递归前序遍历
*
* @param root
* @return
*/
public static List<Integer> preOrderIterator(TreeNode root) {
List<Integer> ans = new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
stack.add(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
if (node != null) {
ans.add(node.val);
//先放入右节点
stack.add(node.right);
stack.add(node.left);
}
}
return ans;
}
/**
* 非递归中序遍历
*
* @param root
* @return
*/
public static List<Integer> midOrderIterator(TreeNode root) {
List<Integer> ans = new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
while (root != null || !stack.empty()) {
//dfs,将所有左节点压栈，然后依次pop出来，达到回溯的效果
while (root != null) {
stack.add(root);
root = root.left;
}
//去除最左的叶子节点
root = stack.pop();
ans.add(root.val);
//把右子树放到栈中
root = root.right;
}
return ans;
}
/**
* 非递归后续遍历
* 解法一：逆序插入值
*
* @param root
* @return
*/
public static List<Integer> postOrderIterator(TreeNode root) {
LinkedList<Integer> ans = new LinkedList<>();
if (root == null) {
return ans;
}
Stack<TreeNode> stack = new Stack<>();
stack.add(root);
while (!stack.empty()) {
TreeNode node = stack.pop();
//先将根节点插入最后一位，然后将左节点右节点推入栈中。
ans.addFirst(node.val);
if (node.left != null) {
stack.push(node.left);
}
if (node.right != null) {
stack.push(node.right);
}
}
return ans;
}
/**
* 非递归后序遍历
* 解法二
*
* @param root
* @return
*/
public static List<Integer> postOrderIterator2(TreeNode root) {
Stack<TreeNode> stack = new Stack<>();
List<Integer> list = new ArrayList<>();
TreeNode pre = null;
TreeNode current = root;
while (current != null || !stack.isEmpty()) {
while (current != null) {
stack.push(current);
current = current.left;
}
current = stack.peek();
//当前节点右子节点不为null且未被访问过
if (current.right != null && pre != current.right) {
current = current.right;
} else {
stack.pop();
list.add(current.val);
pre = current;
current = null;
}
}
return list;
}
/**
* 非递归层次遍历
* (左视图和右视图都是层次遍历的一个变种,取每层最左边的值或者最右边的值)
*
* @param root
* @return
*/
public static List<List<Integer>> leverOrderIterator(TreeNode root) {
List<List<Integer>> ans = new ArrayList<>();
if (root == null) return ans;
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
List<Integer> cur = new ArrayList<>();
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
//放入当前节点
cur.add(node.val);
//将当前节点的左右子节点加入队列中
if (node.left != null) queue.add(node.left);
if (node.right != null) queue.add(node.right);
}
ans.add(cur);
}
return ans;
}
/**
* 层次遍历
*
* @param root
* @return
*/
public static List<List<Integer>> leverOrderRecursion(TreeNode root) {
List<List<Integer>> ans = new ArrayList<>();
leverOrderRecursion(root, ans, 0);
return ans;
}
private static void leverOrderRecursion(TreeNode root, List<List<Integer>> ans, int level) {
if (root == null) return;
if (ans.size() == level) {
ans.add(new ArrayList<>());
}
ans.get(level).add(root.val);
leverOrderRecursion(root.left, ans, level + 1);
leverOrderRecursion(root.right, ans, level + 1);
}
/**
* 之字形层次遍历
* 103. Binary Tree Zigzag Level Order Traversal
* Given binary tree [3,9,20,null,null,15,7],
* 3
* / \
* 9 20
* / \
* 15 7
* return its zigzag level order traversal as:
* [
* [3],
* [20,9],
* [15,7]
* ]
*/
public static List<List<Integer>> levelOrderRecursionZigzag(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
levelOrderRecursionZigzag(root,result,0);
return result;
}
private static void levelOrderRecursionZigzag(TreeNode root, List<List<Integer>> result, int level) {
if(root == null) return;
if(result.size() == level) {
result.add(new LinkedList<>());
}
// 正向遍历
if(level%2 == 0) {
((LinkedList)result.get(level)).add(root.val);
}else {
((LinkedList)result.get(level)).addFirst(root.val);
}
levelOrderRecursionZigzag(root.left, result, level+1);
levelOrderRecursionZigzag(root.right, result, level+1);
}
}