标签:imp bsp des ati lan leave this key lis
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].
Better Solution: https://discuss.leetcode.com/topic/49194/10-lines-simple-java-solution-using-recursion-with-explanation/2
For this question we need to take bottom-up approach. The key is to find the height of each node. The height of a node is the number of edges from the node to the deepest leaf.
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public List<List<Integer>> findLeaves(TreeNode root) { 12 List<List<Integer>> res = new ArrayList<>(); 13 helper(root, res); 14 return res; 15 } 16 17 public int helper(TreeNode cur, List<List<Integer>> res) { 18 if (cur == null) return -1; 19 int level = 1 + Math.max(helper(cur.left, res), helper(cur.right, res)); 20 if (res.size() <= level) 21 res.add(new ArrayList<Integer>()); 22 res.get(level).add(cur.val); 23 cur.left = cur.right = null; 24 return level; 25 } 26 }
First time solution: HashSet+ DFS
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public List<List<Integer>> findLeaves(TreeNode root) { 12 ArrayList<List<Integer>> res = new ArrayList<List<Integer>>(); 13 if (root == null) return res; 14 HashSet<TreeNode> visited = new HashSet<>(); 15 while (!visited.contains(root)) { 16 ArrayList<Integer> leaves = new ArrayList<Integer>(); 17 helper(root, leaves, visited); 18 res.add(new ArrayList<Integer>(leaves)); 19 } 20 return res; 21 } 22 23 public void helper(TreeNode cur, ArrayList<Integer> leaves, HashSet<TreeNode> visited) { 24 if ((cur.left==null || visited.contains(cur.left)) && (cur.right==null || visited.contains(cur.right))) { 25 leaves.add(cur.val); 26 visited.add(cur); 27 return; 28 } 29 if (cur.left!=null && !visited.contains(cur.left)) 30 helper(cur.left, leaves, visited); 31 if (cur.right!=null && !visited.contains(cur.right)) 32 helper(cur.right, leaves, visited); 33 } 34 }
Leetcode: Find Leaves of Binary Tree
标签:imp bsp des ati lan leave this key lis
原文地址:http://www.cnblogs.com/EdwardLiu/p/6193740.html