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Leetcode: Binary Tree Level Order Transversal 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 7return its bottom-up level order traversal as:[ [15,7], [9,20], [3]]
在Binary Tree Level Order Transversal的基础上难度:20,只需要对最后结果做一个倒序就好。格式是Collections.reverse(List<?> list)
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>> lists = new ArrayList<ArrayList<Integer>> ();13 if (root == null) return lists;14 LinkedList<TreeNode> queue = new LinkedList<TreeNode>();15 queue.add(root);16 int ParentNumInQ = 1;17 int ChildNumInQ = 0;18 ArrayList<Integer> list = new ArrayList<Integer>();19 while (!queue.isEmpty()) {20 TreeNode cur = queue.poll();21 list.add(cur.val);22 ParentNumInQ--;23 if (cur.left != null) {24 queue.add(cur.left);25 ChildNumInQ++;26 }27 if (cur.right != null) {28 queue.add(cur.right);29 ChildNumInQ++;30 }31 if (ParentNumInQ == 0) {32 ParentNumInQ = ChildNumInQ;33 ChildNumInQ = 0;34 lists.add(list);35 list = new ArrayList<Integer>();36 }37 }38 Collections.reverse(lists);39 return lists;40 }41 }
Leetcode: Binary Tree Level Order Transversal II
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