剑指 Offer 32 – III. 从上到下打印二叉树
本问题对应的 leetcode 原文链接：剑指 Offer 32 – III. 从上到下打印二叉树
问题描述
请实现一个函数按照之字形顺序打印二叉树，即第一行按照从左到右的顺序打印，第二层按照从右到左的顺序打印，第三行再按照从左到右的顺序打印，其他行以此类推。
例如:
给定二叉树: [3,9,20,null,null,15,7],
3
/
9 20
/
15 7
返回其层次遍历结果：
[
[3],
[20,9],
[15,7]
]
提示：
节点总数 <= 1000
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
public List> levelOrder(TreeNode root) {
if(root == null){
return new ArrayList<>();
}
Queue queue = new LinkedList<>();
List> res = new ArrayList<>();
int sum = 1;
queue.add(root);
while(!queue.isEmpty()){
int k = queue.size();
LinkedList tmp = new LinkedList<>();
for(int i = 0; i < k; i++){
TreeNode t = queue.poll();
if(sum % 2 = = 1){
tmp.add(t.val);
} else {
tmp.addFirst(t.val);
}
if(t.left != null) queue.add(t.left);
if(t.right != null) queue.add(t.right);
}
res.add(tmp);
sum ++;
}
return res;
}
}
时间复杂度：O(n)
额外空间复杂度：容器里最对存放 1/2 的节点，故为 O(n)
Python
# 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 []
queue = [root]
res = []
sum = 1
while queue:
size = len(queue)
tmp = []
for i in range(size):
node = queue.pop(0)
if sum % 2 == 1:
tmp.append(node.val)
else:
tmp.insert(0, node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
res.append(tmp)
sum += 1
return res
C++
/**
* 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> levelOrder(TreeNode* root) {
if (!root) {
return {};
}
queue q{{root}};
vector> res;
int sum = 1;
while (!q.empty()) {
int n = q.size();
vector level(n);
for (int i = 0; i < n; ++i) {
auto t = q.front();
q.pop();
if (sum % 2 == 1) {
level[i] = t->val;
} else {
level[n - 1 - i] = t->val;
}
if (t->left) {
q.push(t->left);
}
if (t->right) {
q.push(t->right);
}
}
res.push_back(level);
++sum;
}
return res;
}
};
Go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func levelOrder(root *TreeNode) [][]int {
if root == nil {
return [][]int{}
}
queue := []*TreeNode{root}
res := [][]int{}
sum := 1
for len(queue) > 0 {
size := len(queue)
tmp := []int{}
for i := 0; i < size; i++ {
node := queue[0]
queue = queue[1:]
if sum % 2 == 1 {
tmp = append(tmp, node.Val)
} else {
tmp = append([]int{node.Val}, tmp...)
}
if node.Left != nil {
queue = append(queue, node.Left)
}
if node.Right != nil {
queue = append(queue, node.Right)
}
}
res = append(res, tmp)
sum++
}
return res
}
JS
/**
* 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) {
if (root == null) {
return [];
}
var queue = [];
var res = [];
var sum = 1;
queue.push(root);
while (queue.length > 0) {
var k = queue.length;
var tmp = [];
for (var i = 0; i < k; i++) {
var t = queue.shift();
if (sum % 2 == 1) {
tmp.push(t.val);
} else {
tmp.unshift(t.val);
}
if (t.left != null) queue.push(t.left);
if (t.right != null) queue.push(t.right);
}
res.push(tmp);
sum++;
}
return res;
};
