Leetcode 366. Find Leaves of Binary Tree

Given a binary tree, collect a tree‘s nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty.

Example:

Given binary tree

          1
         /         2   3
       / \
      4   5

Returns [4, 5, 3], [2], [1].

Explanation:

1. Removing the leaves [4, 5, 3] would result in this tree:

          1
         /
        2

2. Now removing the leaf [2] would result in this tree:

          1

3. Now removing the leaf [1] would result in the empty tree:

          []

Returns [4, 5, 3], [2], [1].

解法一: 一层一层的将所有的叶子拔掉,每次去掉一个叶子,就把这个叶子的val设置成‘#‘。

 1 class Solution(object):
 2     def findLeaves(self, root):
 3         """
 4         :type root: TreeNode
 5         :rtype: List[List[int]]
 6         """
 7         ans = []
 8         if not root:
 9             return ans
10
11         while root.val != ‘#‘:
12             curL = []
13             self.removeCurLeaves(root, curL)
14             ans.append([x for x in curL])
15         return ans
16
17
18     def removeCurLeaves(self, node, res):
19         if (not node.left or node.left.val == ‘#‘) and (not node.right or node.right.val == ‘#‘):
20             res.append(node.val)
21             node.val = ‘#‘
22             return
23         if node.left and node.left.val != ‘#‘:
24             self.removeCurLeaves(node.left, res)
25         if node.right and node.right.val != ‘#‘:
26             self.removeCurLeaves(node.right, res)
时间: 2024-10-17 18:44:24

Leetcode 366. Find Leaves of Binary Tree的相关文章

366. Find Leaves of Binary Tree C#

Example:Given binary tree 1 / 2 3 / \ 4 5 Returns [4, 5, 3], [2], [1]. Explanation: 1. Removing the leaves [4, 5, 3] would result in this tree: 1 / 2 2. Now removing the leaf [2] would result in this tree: 1 3. Now removing the leaf [1] would result

366. Find Leaves of Binary Tree

Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty. Example:Given binary tree 1 / 2 3 / \ 4 5 Returns [4, 5, 3], [2], [1]. Explanation: 1. Removing the leaves [4, 5, 3]

【LeetCode】 Maximum Depth of Binary Tree

Maximum Depth of Binary Tree  Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. 递归基础,不解释. class Solution { public: int getMax(TreeNode *root

leetcode -day24 Maximum Depth of Binary Tree & Binary Tree Zigzag Level Order Traversal

1.Maximum Depth of Binary Tree Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. class Solution { public: int maxDepth(TreeNode *root) { inM

leetcode -day9 Candy & Gas Station & Binary Tree Maximum Path Sum

1.  Candy There are N children standing in a line. Each child is assigned a rating value. You are giving candies to these children subjected to the following requirements: Each child must have at least one candy. Children with a higher rating get m

Leetcode 树 Maximum Depth of Binary Tree

本文为senlie原创,转载请保留此地址:http://blog.csdn.net/zhengsenlie Maximum Depth of Binary Tree Total Accepted: 16605 Total Submissions: 38287 Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the ro

【LeetCode】Minimum Depth of Binary Tree 二叉树的最小深度 java

[LeetCode]Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 递归和非递归,此提比较简单.广度优先遍历即可.关键之处就在于如何保持访问深度. 下面是4种代码: 1

【leetcode刷题笔记】Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / 9 20 / 15 7 return its bottom-up level order traver

【Leetcode长征系列】Construct Binary Tree from Inorder and Postorder Traversal

原题: Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 思路:和上一题一样,后续我们可以通过最后一个值得到根的值,同样可以通过定位根的值得到左右子树的子集,递归求解即可. 代码: /** * Definition for binary tree * struct Tre