Binary Tree Level Order Traversal II

描述

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},

  1.     3
  2.    / \
  3.   9  20
  4.     /  \
  5.    15   7

return its bottom-up level order traversal as:

  1. [
  2.   [15,7]
  3.   [9,20],
  4.   [3],
  5. ]

分析

在上一题 Binary Tree Level Order Traversal 的基础上，reverse()一下即可。

递归版

  1. // Binary Tree Level Order Traversal II
  2. // 递归版，时间复杂度O(n)，空间复杂度O(n)
  3. public class Solution {
  4.     public List<List<Integer>> levelOrderBottom(TreeNode root) {
  5.         List<List<Integer>> result = new ArrayList<>();
  6.         traverse(root, 1, result);
  7.         Collections.reverse(result);
  8.         return result;
  9.     }
  11.     void traverse(TreeNode root, int level, 
  12.                   List<List<Integer>> result) {
  13.         if (root == null) return;
  15.         if (level > result.size())
  16.             result.add(new ArrayList<>());
  18.         result.get(level-1).add(root.val);
  19.         traverse(root.left, level+1, result);
  20.         traverse(root.right, level+1, result);
  21.     }
  22. }

迭代版

  1. // Binary Tree Level Order Traversal II
  2. // 迭代版，时间复杂度O(n)，空间复杂度O(1)
  3. public class Solution {
  4.     public List<List<Integer>> levelOrderBottom(TreeNode root) {
  5.         List<List<Integer>> result = new ArrayList<>();
  6.         Queue<TreeNode> current = new LinkedList<>();
  7.         Queue<TreeNode> next = new LinkedList<>();
  9.         if(root == null) {
  10.             return result;
  11.         } else {
  12.             current.offer(root);
  13.         }
  15.         while (!current.isEmpty()) {
  16.             ArrayList<Integer> level = new ArrayList<>(); // elments in one level
  17.             while (!current.isEmpty()) {
  18.                 TreeNode node = current.poll();
  19.                 level.add(node.val);
  20.                 if (node.left != null) next.add(node.left);
  21.                 if (node.right != null) next.add(node.right);
  22.             }
  23.             result.add(level);
  24.             // swap
  25.             Queue<TreeNode> tmp = current;
  26.             current = next;
  27.             next = tmp;
  28.         }
  29.         Collections.reverse(result);
  30.         return result;
  31.     }
  32. }

相关题目

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/binary-tree/traversal/binary-tree-level-order-traversal-ii.html