107. Binary Tree Level Order Traversal II - jiejackyzhang/leetcode-note GitHub Wiki

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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

Tree类题目。

按tree的level保存数据，很显然可以采用DFS的方式。 注意点为保存数据时是底层的在先，因此应采用queue的形式，每次在0位置插入新的list。 level对应的list可通过下标res.size()-1-level获取。

/**
 * 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>> res = new ArrayList<List<Integer>>();
        helper(root, 0, res);
        return res;
    }
    
    private void helper(TreeNode node, int level, List<List<Integer>> res) {
        if(node == null) return;
        if(res.size() < level+1) {
            res.add(0, new ArrayList<Integer>());
        } 
        List<Integer> list = res.get(res.size() - 1 - level);
        list.add(node.val);
        helper(node.left, level+1, res);
        helper(node.right, level+1, res);
    }
}
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