标签:oid turn 返回 get private linked ini bottom pen
题目链接 : https://leetcode-cn.com/problems/binary-tree-level-order-traversal-ii/
给定一个二叉树,返回其节点值自底向上的层次遍历。 (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],
3
/ 9 20
/ 15 7
返回其自底向上的层次遍历为:
[
[15,7],
[9,20],
[3]
]
与上一题层次遍历一样,只不过输出的顺序取反了!
所以只需要从头添加数组就可以了!
思路一: 迭代
思路二: 递归
思路一:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
from collections import deque
if not root: return []
queue = deque()
queue.appendleft(root)
res = []
while queue:
tmp = []
n = len(queue)
for _ in range(n):
node = queue.pop()
tmp.append(node.val)
if node.left:
queue.appendleft(node.left)
if node.right:
queue.appendleft(node.right)
res.insert(0, tmp)
return res
java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> res = new LinkedList<>();
if (root == null) return res;
Deque<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
List<Integer> tmp = new ArrayList<>();
int n = queue.size();
for (int i = 0; i < n; i++) {
TreeNode node = queue.poll();
tmp.add(node.val);
if (node.left != null) queue.add(node.left);
if (node.right != null) queue.add(node.right);
}
res.add(0, tmp);
}
return res;
}
}
思路二:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
res = []
def helper(root, depth):
if not root: return
if depth == len(res):
res.insert(0, [])
res[-(depth+1)].append(root.val)
helper(root.left, depth+1)
helper(root.right, depth+1)
helper(root, 0)
return res
java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root){
List<List<Integer>> res = new LinkedList<>();
helper(res, root, 0);
return res;
}
private void helper(List<List<Integer>> res, TreeNode root, int depth) {
if (root == null) return;
if (res.size() == depth) res.add(0, new ArrayList<>());
res.get(res.size() - depth - 1).add(root.val);
helper(res, root.left, depth + 1);
helper(res, root.right, depth + 1);
}
}
标签:oid turn 返回 get private linked ini bottom pen
原文地址:https://www.cnblogs.com/powercai/p/11104601.html