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【leetcode】Binary Tree Level Order Traversal - IT-Homer 专栏

[日期:2013-04-19] 来源:  作者: [字体: ]

Question :

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

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


Anwser 1 :         

/**
 * 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> > levelOrder(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function

        vector<vector<int>> ret;
        
        if(root == NULL) return ret;
        
        vector<int> vec;
        queue<TreeNode *> Q;
        
        Q.push(root);
        int count = 1;
        
        while(!Q.empty()){
            
            vec.clear();
            int nextCount = 0;      // cal next row count
            for(int i = 0; i < count; i++){     // one row count
                TreeNode *tmp = Q.front();
                Q.pop();
                
                vec.push_back(tmp->val);        // save one row val
                if(tmp->left){
                    Q.push(tmp->left);
                    nextCount++;
                }
                if(tmp->right){
                    Q.push(tmp->right);
                    nextCount++;
                }
            }
            count = nextCount;
            ret.push_back(vec);
        }
        return ret;
    }
};


Anwser 2 :           

/**
 * 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> > levelOrder(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int>> ret;
        if(root == NULL) return ret;
        
        vector<int> vec;
        queue<TreeNode *> Q;
        queue<TreeNode *> Q2;   // extra space
        
        Q.push(root);
        while(!Q.empty()){
            TreeNode *tmp = Q.front();
            Q.pop();
            
            if(tmp != NULL){
                vec.push_back(tmp->val);
                if(tmp->left) Q2.push(tmp->left);
                if(tmp->right) Q2.push(tmp->right);
            }
            
            if(Q.empty()){      // one row end
                ret.push_back(vec);
                vec.clear();
                swap(Q, Q2);
            }
        }
    }
};






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