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], ]
解题思路和level order traversal 一样, 就是最后把输出结果反转一下
一个queue的方法,reverse 结果
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); if(root == null) return result; LinkedList<TreeNode> queue = new LinkedList<TreeNode>(); queue.add(root); int nextLevelCount = 1; while(!queue.isEmpty()){ int curLevel = nextLevelCount; nextLevelCount = 0; ArrayList<Integer> lvl = new ArrayList<Integer>(); for(int i = 0; i < curLevel; i++){ TreeNode tmp = queue.poll(); lvl.add(tmp.val); if(tmp.left != null){ queue.add(tmp.left); nextLevelCount++; } if(tmp.right != null){ queue.add(tmp.right); nextLevelCount++; } } result.add(lvl); } ArrayList<ArrayList<Integer>> reverse = new ArrayList<ArrayList<Integer>>(); for(int i = result.size()-1; i >= 0; i--){ reverse.add(result.get(i)); } return reverse; } }
用 Stack 和 两个queue 的方法
public class Solution { public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); if(root == null) return result; Stack<ArrayList<Integer>> stack = new Stack<ArrayList<Integer>>(); stack = traversal(root); while(!stack.isEmpty()) result.add(stack.pop()); return result; } private Stack<ArrayList<Integer>> traversal(TreeNode root) { LinkedList<TreeNode> queue = new LinkedList<TreeNode>(); Stack<ArrayList<Integer>> result = new Stack<ArrayList<Integer>>(); queue.add(root); while(!queue.isEmpty()) { ArrayList<Integer> tmp = new ArrayList<Integer>(); LinkedList<TreeNode> tmpqueue = new LinkedList<TreeNode>(); while(!queue.isEmpty()) { TreeNode node = queue.poll(); tmp.add(node.val); if(node.left != null) tmpqueue.add(node.left); if(node.right != null) tmpqueue.add(node.right); } result.add(tmp); queue = tmpqueue; } return result; } }