leetCode 107.Binary Tree Level Order Traversal II （二叉树水平序）

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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]]

思路：此题和二叉树水平序类似，求解水平序之后再反转下链表即可。代码如下：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    List<List<Integer>> list;    public List<List<Integer>> levelOrderBottom(TreeNode root) {        list = new ArrayList<List<Integer>>();        //对list进行倒序排序        dfs(0,root);                for(int i = 0; i+i < list.size()-1;i++){            List<Integer> al = list.get(i);            list.set(i,list.get(list.size()-1-i));            list.set(list.size()-1-i,al);        }        return list;    }        private void dfs(int dep,TreeNode root){        if(root == null){            return;        }        List<Integer> al;        if(list.size() > dep){            al = list.get(dep);        }else{            al = new ArrayList<Integer>();            list.add(al);        }                dfs(dep+1,root.left);        al.add(root.val);        dfs(dep+1,root.right);    }}

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