[LeetCode] Binary Tree Level Order Traversal II

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).

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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

解题思路：

这道题题意是从下往上逐层遍历二叉树，将结果存在一个二维向量中。可以先求出该树的高度，确定二维向量第一维的大小，然后深度遍历数，将值插入到相应的位置中即可。下面的方法只需8ms。另外题目中给出的问题：{1，#，2，3}表示按数组的形式存储二叉树，#表示对应的节点为空，下标能够表示两个节点的相互关系。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        int h = getHeight(root);
        vector<vector<int>> result(h, vector<int>());
        traversal(root, 1, h, result);
        return result;
    }

private:

    void traversal(TreeNode* root, int level, int h, vector<vector<int>>& result){
        if(root!=NULL){
            result[h-level].push_back(root->val);
            traversal(root->left, level+1, h, result);
            traversal(root->right, level+1, h, result);
        }
    }
    
    int getHeight(TreeNode* root){
        if(root==NULL){
            return 0;
        }
        return 1+max(getHeight(root->left), getHeight(root->right));
    }
};

[LeetCode] Binary Tree Level Order Traversal II

原文：http://blog.csdn.net/kangrydotnet/article/details/45484545