二叉树层次遍历c语言_[LeetCode] 107. 二叉树的层次遍历 II

题目链接 : https://leetcode-cn.com/problems/binary-tree-level-order-traversal-ii/

题目描述:

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]
]

思路:

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

代码:

思路一:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:from collections import dequeif 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 = Noneclass 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);}
}