BinaryTreeLevelOrderTraversalII

二叉树的层序遍历 II

题目介绍

二叉树的层序遍历 II

给定一个二叉树,返回其节点值自底向上的层序遍历。 (即按从叶子节点所在层到根节点所在的层,逐层从左向右遍历)

例如:
给定二叉树 [3,9,20,null,null,15,7],

1
2
3
4
5
  3
/ \
9 20
/ \
15 7

返回其自底向上的层序遍历为:

1
2
3
4
5
[
[15,7],
[9,20],
[3]
]

题目解法

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package algorithm;

import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

public class BinaryTreeLevelOrderTraversalII {

public static List<List<Integer>> levelOrderBottom(TreeNode root) {

List<List<Integer>> ans = new LinkedList<>();
if (root == null) {
return ans;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);

while (!queue.isEmpty()) {
List<Integer> temp = new LinkedList<>();
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
temp.add(node.val);

if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
ans.add(0, new LinkedList<>(temp));
}

return ans;
}

public static void main(String[] args) {
TreeNode n1 = new TreeNode(3);
TreeNode n2 = new TreeNode(9);
n1.left = n2;
TreeNode n3 = new TreeNode(20);
n1.right = n3;
TreeNode n4 = new TreeNode(15);
n2.left = null;
n2.right = null;
n3.left = n4;
TreeNode n5 = new TreeNode(7);
n3.right = n5;

levelOrderBottom(n1);
}

public static class TreeNode {
int val;
TreeNode left;
TreeNode right;

TreeNode() {
}

TreeNode(int val) {
this.val = val;
}

TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
}

打印:

1

思路:

思路上,其实和z字形很类似。只不过借助于java的api实现头插入。以后可以试试Go或者C的写法。


BinaryTreeLevelOrderTraversalII
https://yangtzeshore.github.io/2021/07/16/BinaryTreeLevelOrderTraversalII/
作者
Chen Peng
发布于
2021年7月16日
许可协议