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Binary Tree Level Order Traversal.cpp
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Binary Tree Level Order Traversal.cpp
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/*
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
Link: http://www.lintcode.com/en/problem/binary-tree-level-order-traversal/
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]
]
Solution: None
Source: https://github.com/kamyu104/LintCode/blob/master/C%2B%2B/binary-tree-level-order-traversal.cpp
*/
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
/**
* @param root: The root of binary tree.
* @return: Level order a list of lists of integer
*/
public:
vector<vector<int>> levelOrder(TreeNode *root) {
// write your code here
vector<vector<int>> output;
queue<TreeNode *> que;
vector<int> cur_level;
// Enqueue when node is not nullptr.
if (root) {
que.emplace(root);
}
int cur_level_cnt = que.size();
while (!que.empty()) {
TreeNode *node = que.front();
que.pop();
--cur_level_cnt;
cur_level.emplace_back(node->val);
// Enqueue the next level.
if (node->left) {
que.emplace(node->left);
}
if (node->right) {
que.emplace(node->right);
}
// Current level has been all visited.
if (cur_level_cnt == 0) {
cur_level_cnt = que.size();
output.emplace_back(move(cur_level)); // move: copy and delete
}
}
return output;
}
};