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107. 二叉树的层序遍历 II

English Version

题目描述

给定一个二叉树，返回其节点值自底向上的层序遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

例如：
给定二叉树 [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回其自底向上的层序遍历为：

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

解法

同 102，最后反转一下结果即可。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
        if root is None:
            return []
        q = [root]
        res = []
        while q:
            size = len(q)
            t = []
            for _ in range(size):
                node = q.pop(0)
                t.append(node.val)
                if node.left is not None:
                    q.append(node.left)
                if node.right is not None:
                    q.append(node.right)
            res.append(t)
        return res[::-1]

Java

/**
 * 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) {
        if (root == null) return Collections.emptyList();
        Deque<TreeNode> q = new ArrayDeque<>();
        List<List<Integer>> res = new ArrayList<>();
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            List<Integer> t = new ArrayList<>();
            while (size-- > 0) {
                TreeNode node = q.poll();
                t.add(node.val);
                if (node.left != null) q.offer(node.left);
                if (node.right != null) q.offer(node.right);
            }
            res.add(0, t);
        }
        return res;
    }
}

JavaScript

/**
 * 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) {
    if (!root) {
        return [];
    }
    let q = [], output = [], levelOutput = [];
    q.push(root);
    q.push(null);
    while (q.length) {
        let cur = q.shift();
        levelOutput.push(cur.val);
        if (cur.left) {
            q.push(cur.left);
        }
        if (cur.right) {
            q.push(cur.right);
        }
        if (!q[0]) {
            q.shift();
            output.unshift(levelOutput);
            levelOutput = [];
            if (q.length) {
                q.push(null);
            }
        }
    }
    return output;
};

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