leetcode-102-Binary-Tree-Level-Order-Traversal

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描述

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},

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  3
 / \
9  20
  /  \
 15   7

return its level order traversal as:

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

分析

层次遍历，而且顺序正常，按常规思路写即可，可以用递归，也可以用迭代，有意思的是代码与(E) Binary Tree Level Order Traversal II几乎一样，只有最后一行不同。用迭代的方式，时间复杂度是O(n)，空间复杂度是O(n)。

解决方案1(Python)

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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        result = []
        next_level = [root]
        
        while next_level:
            now_level = next_level
            result.append(map(lambda x: x.val, now_level))
            
            next_level = []
            for item in now_level:
                if item.left:
                    next_level.append(item.left)
                if item.right:
                    next_level.append(item.right)
        return result

解决方案2(C++)

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/**
 * 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>> result;
    
    void leveltravel(TreeNode* root, int level) {
        if(root == NULL) {
            return;
        }
        if(level == result.size()) {
            vector<int> v;
            result.push_back(v);
        }
        result[level].push_back(root->val);
        leveltravel(root->left, level+1);
        leveltravel(root->right, level+1);
    }
    
    vector<vector<int>> levelOrder(TreeNode* root) {
        leveltravel(root, 0);
        return vector<vector<int>>(result.begin(), result.end());
    }
};

解决方案3(Java)

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new LinkedList<List<Integer>>();
        DFS(root, 0, result);
        return result;
    }
    
    private void DFS(TreeNode node, int i, List<List<Integer>> result) {
        if (node == null) {
            return;
        }
        
        if (i == result.size()) {
            result.add(result.size(), new LinkedList<Integer>());
        }
        result.get(i).add(node.val);
        DFS(node.left, i+1, result);
        DFS(node.right, i+1, result);
    }
}

解决方案4（golang）

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func levelOrder(root *TreeNode) [][]int {
    result := [][]int{}
    if root == nil {
        return result
    }
    
    queue := []*TreeNode{root}
    for i := 0; len(queue) > 0; i++ {
        result = append(result, []int{})
        nowQueue := []*TreeNode{}
        
        for j := 0; j < len(queue); j++ {
            node := queue[j]
            result[i] = append(result[i], node.Val)
            if node.Left != nil {
                nowQueue = append(nowQueue, node.Left)
            }
            if node.Right != nil {
                nowQueue = append(nowQueue, node.Right)
            }
        }
        queue = nowQueue
    }
    return result
}

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