【LeetCode】107. Binary Tree Level Order Traversal II (2 solutions)

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

 

解法一：递归

参考了Discussion中stellari的做法，递归进行层次遍历，并将每个level对应于相应的vector。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution
{
public:
    vector<vector<int> > result;

    void levelTra(TreeNode *root, int level)
    {
        if(root == NULL)
            return;
        if(level == result.size())
        {
            vector<int> v;
            result.push_back(v);
        }
        result[level].push_back(root->val);
        levelTra(root->left, level+1);
        levelTra(root->right, level+1);
    }

    vector<vector<int> > levelOrderBottom(TreeNode *root) 
    {
        levelTra(root, 0);
        return vector<vector<int> >(result.rbegin(), result.rend());
    }
};

 

解法二：普通层次遍历后逆序。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 
struct Node
{
    TreeNode* tNode;
    int level;
    Node(TreeNode* newtNode, int newlevel): tNode(newtNode), level(newlevel) {}
};

class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        vector<vector<int> > ret;
        if(!root)
            return ret;
        // push root 
        Node* rootNode = new Node(root, 0);
        queue<Node*> Nqueue;
        Nqueue.push(rootNode);
        
        vector<int> cur;
        int curlevel = 0;
        while(!Nqueue.empty())
        {
            Node* frontNode = Nqueue.front();
            Nqueue.pop();
            
            if(frontNode->level > curlevel)
            {
                ret.push_back(cur);
                cur.clear();
                curlevel = frontNode->level;
            }

            cur.push_back(frontNode->tNode->val);
            
            if(frontNode->tNode->left)
            {
                Node* leftNode = new Node(frontNode->tNode->left, frontNode->level+1);
                Nqueue.push(leftNode);
            }
            if(frontNode->tNode->right)
            {
                Node* rightNode = new Node(frontNode->tNode->right, frontNode->level+1);
                Nqueue.push(rightNode);
            }
        }
        ret.push_back(cur);
        
        reverse(ret.begin(), ret.end());
        return ret;
    }
};