leetcode | 107. 二叉树的层序遍历 II | javascript实现 | c++实现

题目

给你二叉树的根节点 root ，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）

思路

题目的要求相当于是求层序遍历数组的转置，我们只需利用js的unshift对返回数组进行头插法即可

代码

JavaScript实现

/**
 * 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 curLayer = 0; // 当前层节点数
    let nextLayer = 0; // 下一层节点数
    let layer = []; // 存储当前层节点列表
    let queue = []; // 队列
    let front; // 队头
    let res = []; // 返回
    if (root) {
        queue.push(root);
        curLayer++;
    }

    while (queue.length) {
        front = queue.shift();
        layer.push(front.val);
        if (front.left) {
            queue.push(front.left);
            nextLayer++;
        }
        if (front.right) {
            queue.push(front.right);
            nextLayer++;
        }

        if (--curLayer === 0) {
            curLayer = nextLayer;
            nextLayer = 0;
            res.unshift(layer);
            layer = [];
        }
    }

    return res;
};

c++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<int> layer;
        queue<TreeNode*> q;
        TreeNode* front;
        vector<vector<int>> res;
        int curLayer = 0;
        int nextLayer = 0;

        if (root) {
            q.push(root);
            curLayer++;
        }
        while (!q.empty()) {
            front = q.front();
            q.pop();
            layer.push_back(front->val);

            if (front->left) {
                q.push(front->left);
                nextLayer++;
            }
            if (front->right) {
                q.push(front->right);
                nextLayer++;
            }

            if (--curLayer == 0) {
                curLayer = nextLayer;
                nextLayer = 0;
                res.push_back(layer);
                layer.clear();
            }
        }
        // reverse
        int idx = res.size() - 1;
        for (int k = 0; k < idx; k += 1, idx--) {
            layer = res[idx];
            res[idx] = res[k];
            res[k] = layer;
        }
        return res;
    }
};