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).
For example:
Given binary tree [3,9,20,null,null,15,7]
,
3
/ 9 20
/ 15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
按从下到上的顺序记录二叉树的每一层。
BFS后再将结果集逆序。
/**
* 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>> levelOrderBottom(TreeNode root) {
List<List<Integer>> ans = new ArrayList<>();
Queue<TreeNode> q = new ArrayDeque<>();
if (root != null) {
q.offer(root);
}
while (!q.isEmpty()) {
List<Integer> list = new ArrayList<>();
int size = q.size();
for (int i = 0; i < size; i++) {
TreeNode cur = q.poll();
list.add(cur.val);
if (cur.left!=null) q.offer(cur.left);
if (cur.right!=null) q.offer(cur.right);
}
ans.add(list);
}
Collections.reverse(ans);
return ans;
}
}
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
var levelOrderBottom = function (root) {
let bfs = []
let q = []
if (root) {
q.push(root)
}
while (q.length > 0) {
let size = q.length
let tmp = []
while (size > 0) {
let p = q.shift()
tmp.push(p.val)
if (p.left) q.push(p.left)
if (p.right) q.push(p.right)
size--
}
bfs.push(tmp)
}
return bfs.reverse()
}
0107. Binary Tree Level Order Traversal II (E)
原文:https://www.cnblogs.com/mapoos/p/13228051.html