Binary Tree Level Order Traversal,Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal

Total Accepted: 79463 Total Submissions: 259292 Difficulty: Easy

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]
]

 (M) Binary Tree Zigzag Level Order Traversal (E) Binary Tree Level Order Traversal II (E) Minimum Depth of Binary Tree (M) Binary Tree Vertical Order Traversal

/**
 * 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;
        vector<int> one_res;

        TreeNode* p = root;
        TreeNode* first = NULL;

        queue<TreeNode*> que;
        if(p) que.push(p);

        while(!que.empty()){
            p = que.front();
            que.pop();

            if(first == p){//碰到每层的第一个时就把上一层次的所有结点加入结果集
                res.push_back(one_res);
                one_res.clear();
                first = NULL;
            }

            one_res.push_back(p->val);

            if(first==NULL && p->left!=NULL){
                first = p->left;
            }
            if(first==NULL && p->right!=NULL){
                first = p->right;
            }

            if(p->left){
                que.push(p->left);
            }
            if(p->right){
                que.push(p->right);
            }
        }

        if(!one_res.empty()){
            res.push_back(one_res);
        }
        return res;
    }
};

Next challenges: (M) Binary Tree Zigzag Level Order Traversal (E) Binary Tree Level Order Traversal II (M) Binary Tree Vertical Order Traversal

 

Binary Tree Level Order Traversal II

Total Accepted: 62827 Total Submissions: 194889 Difficulty: Easy

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]
]

1.正序再反转,8ms

/**
 * 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 {
private:
    void levelOrderBottom(TreeNode* root,vector<vector<int>>& res,int depth){
        if(!root) return;
        if(depth==res.size()){
            res.push_back({});
        }
        res[depth].push_back(root->val);
        levelOrderBottom(root->left,res,depth+);
        levelOrderBottom(root->right,res,depth+);
    }
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> res;
        levelOrderBottom(root,res,);
        reverse(res.begin(),res.end());
        return res;
    }
};

2.先求高度，无需反转，4ms

/**
 * 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 {
private:
    int getTreeHeith(TreeNode* root){
        if(!root) return ;
        return max(getTreeHeith(root->left) ,getTreeHeith(root->right)) +  ;
    }
    void levelOrderBottom(TreeNode* root,vector<vector<int>>& res,int depth){
        if(!root) return;
        res[depth].push_back(root->val);
        levelOrderBottom(root->left,res,depth-);
        levelOrderBottom(root->right,res,depth-);
    }
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        int dep = getTreeHeith(root);
        vector<vector<int>> res(dep,vector<int>());
        levelOrderBottom(root,res,dep-);
        return res;
    }
};