[LeetCode] Binary Tree Level Order Traversal Solution

Given a binary tree, return the level order[/i] 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.
» Solve this problem

[Thoughts]
Two ways to resolve this problem:
1. Breadth first search
Initial an int variable to track the node count in each level and print level by level. And here need a QUEUE as a helper.
2. Depth first search
Rely on the recursion. Decrement level by one as you advance to the next level. When level equals 1, you’ve reached the given level and output them.
The cons is, DFS will revisit the node, which make it less efficient than BFS.

BFS solution
[code]1:    vector<vector<int> > levelOrder(TreeNode *root) {
2:      // Start typing your C/C++ solution below
3:      // DO NOT write int main() function
4:      vector<vector<int> > result;
5:      vector<TreeNode*> sta;
6:      if(root == NULL) return result;
7:      sta.push_back(root);
8:      int nextLevCou=1;
9:      int index=0;
10:      while(index < sta.size())
11:      {
12:        int curLevCou = nextLevCou;
13:        nextLevCou =0;
14:        vector<int> level;
15:        for(int i =index; i< index+curLevCou; i++)
16:        {
17:          root = sta[i];
18:          level.push_back(root->val);
19:          if(root->left!=NULL)
20:          {
21:            sta.push_back(root->left);
22:            nextLevCou++;
23:          }
24:          if(root->right!=NULL)
25:          {
26:            sta.push_back(root->right);
27:            nextLevCou++;
28:          }
29:        }
30:        result.push_back(level);
31:        index = index +curLevCou;
32:      }
33:      return result;
34:    }

DFS solution

1:    vector<vector<int> > levelOrder(TreeNode *root) {
2:      // Start typing your C/C++ solution below
3:      // DO NOT write int main() function
4:      vector<vector<int> > output;
5:      if(!root) return output;
6:      vector<int> oneLine;
7:      bool hasNextLevel=true;
8:      int currentLevel =1;
9:      while(hasNextLevel)
10:      {
11:        hasNextLevel = false;
12:        LevelTravel(root, currentLevel, hasNextLevel, oneLine);
13:        output.push_back(oneLine);
14:        currentLevel ++;
15:        oneLine.clear();
16:      }
17:      return output;
18:    }
19:    void LevelTravel(
20:      TreeNode* node,
21:      int level,
22:      bool& hasNextLevel,
23:      vector<int>& result)
24:    {
25:      if(!node) return;
26:      if(level ==1)
27:      {
28:        result.push_back(node->val);
29:        if(node->left || node->right)
30:          hasNextLevel = true;
31:        return;
32:      }
33:      else
34:      {
35:        LevelTravel(node->left, level-1, hasNextLevel, result);
36:        LevelTravel(node->right, level-1, hasNextLevel, result);
37:      }
38:    }

Update 1/12/2014
BFS的code太啰嗦，用两个循环虽然看着清楚了，但是code不够漂亮，改一下。

1:    vector<vector<int> > levelOrder(TreeNode *root) {
2:      vector<vector<int> > result;
3:      if(root == NULL) return result;
4:      queue<TreeNode*> nodeQ;
5:      nodeQ.push(root);
6:      int nextLevelCnt=0, currentLevelCnt=1;
7:      vector<int> layer;
8:      int visitedCnt=0;
9:      while(nodeQ.size() != 0)
10:      {
11:        TreeNode* node = nodeQ.front();
12:        nodeQ.pop();
13:        visitedCnt++;
14:        layer.push_back(node->val);
15:        if(node->left != NULL)
16:        {
17:          nodeQ.push(node->left);
18:          nextLevelCnt++;
19:        }
20:        if(node->right != NULL)
21:        {
22:          nodeQ.push(node->right);
23:          nextLevelCnt++;
24:        }
25:        if(visitedCnt == currentLevelCnt)
26:        {
27:          visitedCnt =0;
28:          currentLevelCnt = nextLevelCnt;
29:          nextLevelCnt=0;
30:          result.push_back(layer);
31:          layer.clear();
32:        }
33:      }
34:      return result;
35:    }