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107 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]
]

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 
 //方法一：BFS
struct Node
{
  TreeNode *node;
  int level;
  Node(TreeNode *n , int l) : node(n) , level(l) {}
};

class Solution {
vector<vector<int>> ans;
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        ans.clear();
        if(root == NULL)
            return ans;
        
        vector<int> temp;
        int Curlevel = -1;
        
        queue<Node> q;
        q.push( Node(root , 1) );
        
        while(!q.empty())
        {
            Node nodeQ = q.front();
            q.pop();
            
            if(nodeQ.node->left)
                q.push( Node(nodeQ.node->left , nodeQ.level+1) );
            if(nodeQ.node->right)
                q.push( Node(nodeQ.node->right , nodeQ.level+1) );
            
            if(Curlevel != nodeQ.level)
            {
                if(Curlevel != -1)
                    ans.push_back(temp);
                
                temp.clear();
                temp.push_back(nodeQ.node->val);
                Curlevel = nodeQ.level;
            }
            else
                temp.push_back(nodeQ.node->val);
        }
        ans.push_back(temp);
        
        reverse(ans.begin() , ans.end());
        return ans;
    }
};

 //方法二：DFS
class Solution {

vector<vector<int>> ans;
public:
    void levelOrderUtil (TreeNode *root , int depth)
    {
        if(root==NULL)
            return;
        if(ans.size() > depth)
            ans[depth].push_back(root->val);
        else
        {
            vector<int> a;
            a.push_back(root->val);
            ans.push_back(a);
        }
        levelOrderUtil( root->left , depth+1 );
        levelOrderUtil( root->right , depth+1 );
    }
    
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        ans.clear();
        levelOrderUtil( root , 0);
        reverse(ans.begin(), ans.end());
        return ans;
    }
};

leetcode_107_Binary Tree Level Order Traversal II

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原文地址：http://blog.csdn.net/keyyuanxin/article/details/44216829