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Binary Tree Level Order Traversal

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.

解题思路：

用一个队列记录所有的节点。但是这里有个陷阱就是每一层应该分别用一个vector表示，因此还需要一个队列来存储对应的node所在的层数。因为同一层在队列中肯定是相邻的，因此可以通过这种方法来解决。

/**
 * 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>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        if(root==NULL){
            return result;
        }
        queue<int> level;
        queue<TreeNode*> queues;
        level.push(0);
        queues.push(root);
        int lastLevel = 0;
        vector<int> lastValue;
        while(!queues.empty()){
            TreeNode* node = queues.front();
            int l = level.front();
            if(l!=lastLevel){
                result.push_back(lastValue);
                lastValue.clear();
                lastLevel = l;
            }
            lastValue.push_back(node->val);
            queues.pop();
            level.pop();
            if(node->left!=NULL){
                queues.push(node->left);
                level.push(l + 1);
            }
            if(node->right!=NULL){
                queues.push(node->right);
                level.push(l + 1);
            }
        }
        if(!lastValue.empty()){
            result.push_back(lastValue);
        }
        return result;
    }
};

二次刷题（2015-08-03）

用两个队列来存储中间结果。

/**
 * 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>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        if(root==NULL){
            return result;
        }
        queue<TreeNode*> q[2];
        int k = 0;
        q[0].push(root);
        while(!q[k].empty()){
            vector<int> item;
            while(!q[k].empty()){
                TreeNode* node = q[k].front();
                q[k].pop();
                item.push_back(node->val);
                if(node->left!=NULL){
                    q[(k+1)%2].push(node->left);
                }
                if(node->right!=NULL){
                    q[(k+1)%2].push(node->right);
                }
            }
            k = (k+1)%2;
            result.push_back(item);
        }
        
        return result;
    }
};

用深度优先遍历来实现

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