LeetCode *** 102. Binary Tree Level Order Traversal-优快云博客

本文介绍了一种使用广度优先搜索(BFS)算法来实现二叉树的层次遍历的方法，并通过C++代码示例详细展示了如何从左到右逐层获取二叉树节点的值。

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题目：

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},

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

分析：

广度优先搜索

代码：

/**
 * 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>> res;
        
        queue<TreeNode*> tree;
        queue<int> level;
        
        if(root!=NULL){
            tree.push(root);
            level.push(0);
        }
        
        vector<int> resNow;
        
        while(!tree.empty()){
            TreeNode* now=tree.front();
            int levelNow=level.front();
            tree.pop();
            level.pop();
            
            resNow.insert(resNow.end(),now->val);
            
            if(tree.empty()||level.front()!=levelNow){
                res.insert(res.end(),resNow);
                resNow.erase(resNow.begin(),resNow.end());
            }
            
            if(now->left){
                tree.push(now->left);
                level.push(levelNow+1);
            }
            if(now->right){
                tree.push(now->right);
                level.push(levelNow+1);
            }
        }
        
        return res;
    }
};