leetcode：Binary Tree Level Order Traversal II 【Java】

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本文介绍了一种从叶到根的层次遍历二叉树的方法，并提供了详细的算法实现步骤。通过对二叉树进行层次遍历，再反转结果，实现了从叶子节点到根节点的遍历。

一、问题描述

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},

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

二、问题分析

利用leetcode：Binary Tree Level Order Traversal遍历该树，然后翻转结果。

三、算法代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if(root == null){
            return result;
        }
        List<TreeNode> pre = new ArrayList<TreeNode>();//保存树中pre层树结点
        pre.add(root);
        int preLength = 0;
        List<TreeNode> cur = null;//保存树中当前层树结点
        List<Integer> curValue = null;//保存树的当前层树节点中数值
        TreeNode curNode = null;
        while(pre.size() != 0){
            preLength = pre.size();
            cur = new ArrayList<TreeNode>();
            curValue = new ArrayList<Integer>();
            for(int i = 0; i <= preLength - 1; i++){
                curNode = pre.get(i);
                curValue.add(curNode.val);//保存树的当前层树节点中数值
                if(curNode.left != null){
                    cur.add(curNode.left); 
                }
                if(curNode.right != null){
                    cur.add(curNode.right); 
                }
            }
            result.add(curValue);
            pre = cur;
        }
        
        //reverse result
        List<Integer> levelValue = null;
        int front = 0;
        int tail = result.size() - 1;
        while(front <= tail){
            levelValue = result.get(front);
            result.set(front, result.get(tail));
            result.set(tail, levelValue);
            front++;
            tail--;
        }
        return result;
    }
}