leetcode-tree-107-Binary Tree Level Order Traversal II

107 Binary Tree Level Order Traversal II

题目描述

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).

给定二叉树，返回其自下而上的遍历顺序的节点值

• 输入

    3
   / \
  9  20
    /  \
   15   7

• 输出

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

解法

let levelOrderBottom = function(root) {
    if (!root) { return []} // 不能少，不然lever.push报错
    let result = [] // 用来存放最终结果
    let queue = [root] 
    while(queue.length){ 
        let level = []; // 存放每一层的元素
        let len = queue.length; 
       
        for(let i = 0; i < len; i++){ // 通过遍历将当前层所有元素放进level
            let node = queue.shift()
            level.push(node.val) 

            if(node.left){queue.push(node.left) } // 如果该节点下一层还有值，则放入队列中，一下遍历继续取出
            if(node.right) {queue.push(node.right)}
        }

        if(level.length) {result.push(level)}

    }
    return result.reverse() // 这样取出来的值从上往下一层一层取，结果要的是从底部网上
};

按照上面例子就是
1 queue = [3]
2 node = 3, level = [3], queue = []
3 node.left = 9, node.right = 20, queue = [9,20]

下一轮循环
1 queue = [9,20]
2 node = 9,level = [9],queue = [20]
3 node = 20,level = [9,20], queue = []
4 node.left = 15, node.right = 7,queue = [17,7]

再一次下一轮。。。