Binary Tree Level Order Traversal

Source

• lintcode: (69) 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).

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]
]
Challenge
Using only 1 queue to implement it.

题解 - 使用队列

此题为广搜的基础题，使用一个队列保存每层的节点即可。

C++ queue

/**
 * 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) {
        vector<vector<int> > result;

        if (NULL == root) {
            return result;
        }

        queue<TreeNode *> q;
        q.push(root);
        while (!q.empty()) {
            vector<int> list;
            int size = q.size(); // keep the queue size first
            for (int i = 0; i != size; ++i) {
                TreeNode * node = q.front();
                q.pop();
                list.push_back(node->val);
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
            result.push_back(list);
        }

        return result;
    }
};

源码分析

1. 异常，还是异常
2. 使用STL的queue数据结构，将root添加进队列
3. 遍历当前层所有节点，注意需要先保存队列大小，因为在入队出队时队列大小会变化
4. list保存每层节点的值，每次使用均要初始化