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] ]
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], ]
思路：
把栈式的访问改成队列式的访问。深度优先搜索改为广度优先搜索。
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) {
if (!root)
return vector<vector<int>>();
queue<pair<TreeNode*, int>> visit_list;
vector<vector<int>> ret;
visit_list.push(make_pair(root, 0));
while(!visit_list.empty())
{
while(visit_list.front().second >= ret.size())
ret.push_back(vector<int>());
ret[visit_list.front().second].push_back(visit_list.front().first->val);
if (visit_list.front().first->left != nullptr)
visit_list.push(make_pair(visit_list.front().first->left, visit_list.front().second + 1));
if (visit_list.front().first->right != nullptr)
visit_list.push(make_pair(visit_list.front().first->right, visit_list.front().second + 1));
visit_list.pop();
}
return ret;
}
};
思路：
同上，只不过多一次std::reverse的工作。注意这里用reverse并不会带来vector的值交换，而只是交换vector内部的指针而已，所以性能没有什么损失。
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>> retval;
if (root == nullptr)
return retval;
queue<pair<TreeNode*, int>> traverse;
traverse.push(make_pair(root, 1));
while(!traverse.empty())
{
auto front = traverse.front();
traverse.pop();
if (retval.size() < front.second)
retval.push_back(vector<int>());
retval.back().push_back(front.first->val);
if (front.first->left != nullptr)
traverse.push(make_pair(front.first->left, front.second + 1));
if (front.first->right != nullptr)
traverse.push(make_pair(front.first->right, front.second + 1));
}
// since std::swap is used
// there is actually no vector swap, just pointers swap
reverse(begin(retval), end(retval));
return retval;
}
};