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Binary Tree Level Order Traversa

Binary Tree Level Order Traversa

作者: sunner168 | 来源:发表于2017-08-13 23:18 被阅读7次

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,null,null,15,7],

  3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

python 答案

# 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]]
        """
        result = []
        if not root:
            return result
        queue = [root]
        while queue:
            new_q = []
            result.append([x.val for x in queue])
            for node in queue:
                if node.left:
                    new_q.append(node.left)
                if node.right:
                    new_q.append(node.right)
            queue = new_q
        return result

js答案

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrder = function(root) {
    var result =[],queue = [];
    if(!root){
        return result
    }
    queue.push(root);
    while(queue.length > 0){
       var currentLevel = [ ];
        //current
        var size = queue.length;
        for(var i = 0; i<size; i++){
            
            var element = queue.shift();
             currentLevel.push(element.val)
            if(element.left){
                queue.push(element.left)
            }
           
            if(element.right){
                queue.push(element.right)
            }
            
        }
        result.push(currentLevel)
    }
    return result
};

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