Given a rooted binary tree, return the lowest common ancestor of its deepest leaves. Recall that: The node of a binary tree is a leaf if and only if it has no children The depth of the root of the tree is 0, and if the depth of a node is d, the depth…
[129-Sum Root to Leaf Numbers(全部根到叶子结点组组成的数字相加)] [LeetCode-面试算法经典-Java实现][全部题目文件夹索引] 原题 Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. An example is the root-to-leaf path 1->2->3 which represent…
［抄题］: Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty. Example:Given binary tree 1 / \ 2 3 / \ 4 5 Returns [4, 5, 3], [2], [1]. Explanation: 1. Removing the leaves […
#include <iostream> using namespace std; struct Tree { int data; Tree *lchild; Tree *rchild; }tree; Tree *Create(int a1[],int b1[],int n) { int k; ) return NULL; ]; Tree *bt=(Tree *)malloc(sizeof(Tree)); bt->data=root; ;k<n;k++) { if(b1[k]==ro…
Given a binary tree rooted at root, the depth of each node is the shortest distance to the root. A node is deepest if it has the largest depth possible among any node in the entire tree. The subtree of a node is that node, plus the set of all descend…
Given a binary tree rooted at root, the depth of each node is the shortest distance to the root. A node is deepest if it has the largest depth possible among any node in the entire tree. The subtree of a node is that node, plus the set of all descend…
题目 404. 左叶子之和 如题 题解 类似树的遍历的递归 注意一定要是叶子结点 代码 class Solution { public int sumOfLeftLeaves(TreeNode root) { if(root == null){return 0;} int sum = sumOfLeftLeaves(root.left)+sumOfLeftLeaves(root.right); if(root.left!=null&&root.left.left==null&&am…
刷题备忘录,for bug-free leetcode 396. Rotate Function 题意: Given an array of integers A and let n to be its length. Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow: F(k…
463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…
前言 之前把一些LeetCode题目的思路写在了本子上,现在把这些全都放到博客上,以后翻阅比较方便. 题目 99.Recover Binary Search Tree 题意 Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. 恢复两个因错误而导致的左右子树. 思路 通常我们的做法是,根据BST的特质,可以用中序遍历来检…
