标签：level evel rem count nbsp this binary eve pre
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].
Solution:
Mark tree by level , if it is leave then mark as level 0, then add to the List<List<int>> by level.
/** * Definition for a binary tree node. * public class TreeNode { * public int val; * public TreeNode left; * public TreeNode right; * public TreeNode(int x) { val = x; } * } */ public class Solution { public IList<IList<int>> FindLeaves(TreeNode root) { IList<IList<int>> result = new List<IList<int>> (); LeavesLevel(root, result); return result; } public int LeavesLevel(TreeNode root, IList<IList<int>> result) { if(root==null) { return -1; } int leftLevel = LeavesLevel(root.left, result); int rightLevel = LeavesLevel(root.right, result); int level = Math.Max(leftLevel, rightLevel)+1; if(result.Count()<level+1) { result.Add(new List<int>()); } result[level].Add(root.val); return level; } }
366. Find Leaves of Binary Tree C#
标签：level evel rem count nbsp this binary eve pre
原文地址：http://www.cnblogs.com/MiaBlog/p/6280446.html
