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Given a binary tree, find all leaves and then remove those leaves. Then repeat the previous steps until the tree is empty.
Example:
Given binary tree
1
/ \
2 3
/ \
4 5
Returns [4, 5, 3], [2], [1].
Explanation:
1. Remove the leaves [4, 5, 3] from the tree
1
/
2
2. Remove the leaf [2] from the tree
1
3. Remove the leaf [1] from the tree
[]
Returns [4, 5, 3], [2], [1].
该题的重点是对tree进行剪枝活动时候，我们可以利用x = change(x) 在recursive函数中。例如， root.left = removeLeaves(root.left, result); root.right = removeLeaves(root.right, result), 这样下层内嵌函数返回的null值可以赋给上层调用函数的root.left/right。 因此递归函数的返回类型很自然应该设为TreeNode
1 public class Solution { 2 private TreeNode removeLeaves(TreeNode root, List<Integer> result) 3 { 4 if(root==null) return null; 5 if((root.left == null)&&(root.right==null)){ 6 result.add(root.val); 7 return null; 8 } 9 10 root.left = removeLeaves(root.left, result); 11 root.right = removeLeaves(root.right, result); 12 13 return root; 14 15 } 16 17 public List<List<Integer>> findLeaves(TreeNode root) { 18 List<List<Integer>> res = new ArrayList<List<Integer>>(); 19 if (root==null) return res; 20 while(root!=null){ 21 List<Integer> leaves = new ArrayList<Integer>(); 22 root = removeLeaves(root, leaves); 23 res.add(leaves); 24 } 25 26 return res; 27 } 28 }
1
3. Remove the leaf [1] from the tree
[]
Returns [4, 5, 3], [2], [1].
Find Leaves of Binary Tree 找二叉树的叶节点
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原文地址：http://www.cnblogs.com/unlockleetcode/p/5689985.html
