给你一个二叉树，请你返回其按 层序遍历 得到的节点值。 （即逐层地，从左到右访问所有节点）
https://leetcode-cn.com/problems/binary-tree-level-order-traversal/
示例1：
二叉树：[3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7[
[3],
[9,20],
[15,7]
]
Java解法
思路：
- 层序边历，通过栈缓存当前数据，因为先进后出，所以先存储右节点，
- 优化使用队列来存储，但效率没有太多提升
package sj.shimmer.algorithm.m3_2021;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.List;
import java.util.Queue;
import java.util.Stack;
import sj.shimmer.algorithm.TreeNode;
/**
* Created by SJ on 2021/3/18.
*/
class D52 {
public static void main(String[] args) {
System.out.println(levelOrder(TreeNode.getInstance(new Integer[]{3, 9, 20, null, null, 15, 7})));
System.out.println(levelOrder(TreeNode.getInstance(new Integer[]{1, 2, 3, 4, null, null, 5})));
System.out.println(levelOrder(TreeNode.getInstance(new Integer[]{3, 9, 20, null, null, 15, 7})));
}
public static List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> results = new ArrayList<>();
Queue<TreeNode> queue = new ArrayDeque<>();
if (root != null) {
queue.add(root);
}
while (!queue.isEmpty()) {
List<Integer> tempList = new ArrayList<>();
Queue<TreeNode> temp = new ArrayDeque<>();
while (!queue.isEmpty()) {
TreeNode poll = queue.poll();
tempList.add(poll.val);
if (poll.left != null) {
temp.add(poll.left);
}
if (poll.right != null) {
temp.add(poll.right);
}
}
results.add(tempList);
queue = temp;
}
return results;
}
public static List<List<Integer>> levelOrder1(TreeNode root) {
List<List<Integer>> results = new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
if (root != null) {
stack.add(root);
}
while (!stack.isEmpty()){
List<Integer> tempList = new ArrayList<>();
Stack<TreeNode> temp = new Stack<>();
while (!stack.isEmpty()) {
TreeNode pop = stack.pop();
if (pop != null) {
tempList.add(pop.val);
temp.add(pop.left);
temp.add(pop.right);
}
}
if (tempList.size()!=0) {
results.add(tempList);
while (!temp.isEmpty()) {
stack.add(temp.pop());
}
}
}
return results;
}
}
image
image
官方解
-
广度优先搜索
相较我的处理，优化了临时存储
时间复杂度：O(n)
空间复杂度：O(n)
