# 剑指 Offer 32 - II. 从上到下打印二叉树 II
# 描述
从上到下按层打印二叉树,同一层的节点按从左到右的顺序打印,每一层打印到一行。
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],
[9,20],
[15,7]
]
链接:https://leetcode.cn/problems/cong-shang-dao-xia-da-yin-er-cha-shu-ii-lcof
# 解答
/**
* 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) {
return solution2(root);
};
/*
每次迭代,
当前 queue 的长度就是当前层的节点个数(currentLevelSize)
依次遍历并出队 从 0 到 currentLevelSize 的节点
*/
function solution2(root) {
const queue = [];
const res = [];
if (!root) {
return [];
}
queue.push(root);
while(queue.length > 0) {
const currentLevelRes = [];
const currentLevelSize = queue.length;
res.push(currentLevelRes);
for (let i = 0, len = currentLevelSize; i < len; i++) {
const node = queue.shift();
const { val, left, right } = node;
currentLevelRes.push(val);
if (left) {
queue.push(left);
}
if (right) {
queue.push(right);
}
}
}
return res;
}
/*
给每个 node 添加当前节点的层数(index)
node 的子节点的层数为 index + 1
*/
function solution1(root) {
const queue = [];
const res = [];
if (!root) {
return [];
}
root.index = 0;
queue.push(root);
while(queue.length > 0) {
const node = queue.shift();
const { val, left, right, index } = node;
if (!res[index]) {
res[index] = [];
}
res[index].push(val);
if (left) {
left.index = index + 1;
queue.push(left);
}
if (right) {
right.index = index + 1;
queue.push(right);
}
}
return res;
};
上一篇: 下一篇:
本章目录