【leetcode】Binary Tree Level Order Traversal I & II

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]
]

分层遍历二叉树，用BFS即可

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<TreeNode*> Q;
        vector<vector<int>> ans;
        Q.push(root);
        while(!Q.empty())
        {
            int e = Q.size(); //当前层的结束位置
            vector<int> partans;
            for(int i = ; i < e; i++)
            {
                TreeNode * p = Q.front();
                Q.pop();
                if(NULL != p)
                {
                    partans.push_back(p->val);
                    Q.push(p->left);
                    Q.push(p->right);
                }
            }
            if(!partans.empty())
                ans.push_back(partans);
        }
        return ans;
    }
};

网上DFS的代码：

class Solution {
protected:
    vector<vector<int>> ans;
    void dfs(TreeNode *root, int height){
        if (root == NULL)
            return;
        while (ans.size() <= height)
            ans.push_back(vector<int>());
        ans[height].push_back(root->val);
        dfs(root->left, height + );
        dfs(root->right, height + );
    }

public:
    vector<vector<int>> levelOrder(TreeNode* root) {
       dfs(root, );
        return ans;
    }
};

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]
]

跟上面只是顺序变了，加个reverse就行了。