Binary Tree Level Order Traversal I II

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]
]

思路：

这两个题目都很基础，BFS搜索。

题解：

/**
 * 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) {
        vector<vector<int> > res;
        vector<int> tmp;
        queue<TreeNode *> q;
        int count = 0, num = 1;
        if(root==NULL)
            return res;
        q.push(root);
        while(!q.empty()) {
            for(int i=0;i<num;i++) {
                root = q.front();
                q.pop();
                tmp.push_back(root->val);
                if(root->left) {
                    count++;
                    q.push(root->left);
                }
                if(root->right) {
                    count++;
                    q.push(root->right);
                }
            }
            res.push_back(tmp);
            num = count;
            count = 0;
            tmp.clear();
        }
        return res;
    }
};

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]
]
题解：

/**
 * 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> > levelOrderBottom(TreeNode *root) {
        vector<vector<int> > res;
        vector<int> tmp;
        queue<TreeNode *> q;
        int count = 0, num = 1;
        if(root==NULL)
            return res;
        q.push(root);
        while(!q.empty()) {
            for(int i=0;i<num;i++) {
                root = q.front();
                q.pop();
                tmp.push_back(root->val);
                if(root->left) {
                    count++;
                    q.push(root->left);
                }
                if(root->right) {
                    count++;
                    q.push(root->right);
                }
            }
            res.push_back(tmp);
            num = count;
            count = 0;
            tmp.clear();
        }
        reverse(res.begin(), res.end());
        return res;
    }
};