Question

Source

• leetcode: Binary Tree Level Order Traversal II | LeetCode OJ
• lintcode: (70) 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).

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]
]

题解

此题在普通的 BFS 基础上增加了逆序输出，简单的实现可以使用辅助栈或者最后对结果逆序。

Java - Stack

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left, right;
 *   public TreeNode(int val) {
 *     this.val = val;
 *     this.left = this.right = null;
 *   }
 * }
 */


public class Solution {
  /**
   * @param root: The root of binary tree.
   * @return: buttom-up level order a list of lists of integer
   */
  public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
    ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
    if (root == null) return result;

    Stack<ArrayList<Integer>> s = new Stack<ArrayList<Integer>>();
    Queue<TreeNode> q = new LinkedList<TreeNode>();
    q.offer(root);
    while (!q.isEmpty()) {
      int qLen = q.size();
      ArrayList<Integer> aList = new ArrayList<Integer>();
      for (int i = 0; i < qLen; i++) {
        TreeNode node = q.poll();
        aList.add(node.val);
        if (node.left != null) q.offer(node.left);
        if (node.right != null) q.offer(node.right);
      }
      s.push(aList);
    }

    while (!s.empty()) {
      result.add(s.pop());
    }
    return result;
  }
}

Java - Reverse

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left, right;
 *   public TreeNode(int val) {
 *     this.val = val;
 *     this.left = this.right = null;
 *   }
 * }
 */


public class Solution {
  /**
   * @param root: The root of binary tree.
   * @return: buttom-up level order a list of lists of integer
   */
  public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
    ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
    if (root == null) return result;

    Queue<TreeNode> q = new LinkedList<TreeNode>();
    q.offer(root);
    while (!q.isEmpty()) {
      int qLen = q.size();
      ArrayList<Integer> aList = new ArrayList<Integer>();
      for (int i = 0; i < qLen; i++) {
        TreeNode node = q.poll();
        aList.add(node.val);
        if (node.left != null) q.offer(node.left);
        if (node.right != null) q.offer(node.right);
      }
      result.add(aList);
    }

    Collections.reverse(result);
    return result;
  }
}

源码分析

Java 中 Queue 是接口，通常可用 LinkedList 实例化。

复杂度分析

时间复杂度为 O(n)O(n)O(n), 使用了队列或者辅助栈作为辅助空间，空间复杂度为 O(n)O(n)O(n).