Binary Tree Level Order Traversal II
Link: http://oj.leetcode.com/problems/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 traversal as:
[ [15,7] [9,20], [3], ]
1 /** 2 * Definition for binary tree 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 ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) { 12 ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); 13 if (root == null) 14 return result; 15 // the algo is based on bfs, here I need a queue data structure 16 ArrayList<TreeNode> queue = new ArrayList<TreeNode>(); 17 queue.add(root); 18 ArrayList<Integer> root_value = new ArrayList<Integer>(); 19 // this is a special case, the value of the root should 20 // be added first. 21 root_value.add(root.val); 22 result.add(root_value);// 23 while (!queue.isEmpty()) { 24 // all the nodes of the next level 25 ArrayList<TreeNode> next_level = new ArrayList<TreeNode>(); 26 for (TreeNode t : queue) { 27 // iterate through the queue 28 ArrayList<TreeNode> temp_treenode = getChildren(t); 29 // append all of the next level node to next_level 30 next_level.addAll(temp_treenode); 31 } 32 // clear the queue 33 queue = new ArrayList<>(); 34 // special case, if there's no next level nodes 35 if (next_level.size() != 0) { 36 queue.addAll(next_level); 37 result.add(getValue(next_level)); 38 } 39 } 40 ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>(); 41 for(int i = result.size()-1;i>=0;i--){ 42 res.add(result.get(i)); 43 } 44 return res; 45 } 46 public ArrayList<Integer> getValue(ArrayList<TreeNode> list) { 47 ArrayList<Integer> result = new ArrayList<Integer>(); 48 for (TreeNode t : list) { 49 result.add(t.val); 50 } 51 return result; 52 } 53 54 public ArrayList<TreeNode> getChildren(TreeNode root) { 55 ArrayList<TreeNode> result = new ArrayList<TreeNode>(); 56 if (root.left != null) 57 result.add(root.left); 58 if (root.right != null) 59 result.add(root.right); 60 return result; 61 } 62 }
跟I差不多。revere一下I的结果。。
