Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 简单题,不过注意叶子结点是两个子树都是NULL的指针.只有一个子树NULL的不是. struct TreeNode { int val; TreeNode *left; TreeN…

Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 Trivia:This problem was inspired by this original tweet by Max Howell: Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree…

Description Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. Example Given binary tr…

题目: Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. 代码:…

1.直接把递归把左右子树翻转即可 AC代码: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void invert(TreeNode* root) {…

原题 寻找二叉树最短深度 这里用了dfs,beat 100%,4ms class Solution { public: int minDepth(TreeNode *root, int minNum = 0) { if (root == NULL) { return minNum == 0 ? minNum : 999999; } if (root->left == NULL && root->right == NULL) return minNum + 1; return m…

原题 思路: 题目其实就是求左右最长深度的和 class Solution { private: int res = 0; public: int diameterOfBinaryTree(TreeNode *root) { dfs(root); return res; } int dfs(TreeNode *root) { if (root == NULL) { return 0; } int leftNum = dfs(root->left); int rightNum = dfs(root…

原题链接 水题 深度搜索每一节点的左右深度,左右深度差大于1就返回false. class Solution { public: bool isBalanced(TreeNode *root) { bool flag = true; if (!root) return true; dfs(root, flag); return flag; } private: int dfs(TreeNode *root, bool &flag) { if (!root) return 0; int leftH…

104. Maximum Depth of Binary Tree -- Easy 方法 使用递归 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int…

226. Invert Binary Tree Easy Invert a binary tree. Example: Input: 4 / \ 2 7 / \ / \ 1 3 6 9 Output: 4 / \ 7 2 / \ / \ 9 6 3 1 Trivia:This problem was inspired by this original tweet by Max Howell: Google: 90% of our engineers use the software you wr…

111. Minimum Depth of Binary Tree Easy Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. Note: A leaf is a node with no children. Example: G…

110. Balanced Binary Tree Easy Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the depth of the two subtrees of every node never differ by more than 1. Exa…

104. Maximum Depth of Binary Tree Easy Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. Note: A leaf is a node with no children. Example: G…

114. Flatten Binary Tree to Linked List (Medium) 453. Flatten Binary Tree to Linked List (Easy) 解法1: 用stack. class Solution { public: void flatten(TreeNode *root) { if(!root) return; TreeNode *pnode = NULL; stack<TreeNode*> s; s.push(root); while(!s…

606. Construct String from Binary Tree [easy] You need to construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way. The null node needs to be represented by empty parenthesis pair "()". And you…

Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 有序数组变二叉平衡搜索树,不难,递归就行.每次先序建立根节点(取最中间的数),然后用子区间划分左右子树. 一次就AC了 注意:new 结构体的时候对于 struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x)…

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 思想:递归代码如下: /** * Definition for a binary tree node. * public class…

Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 Trivia:This problem was inspired by this original tweet by Max Howell: Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree…

Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two or zero sub-node. If the node has two sub-nodes, then this node's value is the smaller value among its two sub-nodes.…

http://blog.csdn.net/crazy1235/article/details/51474128 花样做二叉树的题……居然还是不会么…… /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ cl…

按层输出二叉树,广度优先. 3 / \ 9 20 / \ 15 7 [ [15,7], [9,20], [3] ] /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ //广度优先遍历 class Solut…

求二叉树的最小深度: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int minDepth(TreeNode* root) { ; int l = m…

求二叉树的最大深度 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int maxDepth(TreeNode* root) { int l,r; ; l…

判断二叉树是否平衡 a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. 以下解法为什么时间复杂度为O(n)? /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNod…

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* invertTree(TreeNode* root) { if (root == NU…

要求:此题正好和Maximum Depth of Binary Tree一题是相反的,即寻找二叉树的最小的深度值:从根节点到最近的叶子节点的距离. 结题思路:和找最大距离不同之处在于:找最小距离要注意(l<r)? l+1:r+1的区别应用,因为可能存在左右子树为空的情况,此时值就为0,但显然值是不为0的(只有当二叉树为空才为0),所以,在这里注意一下即可! 代码如下: struct TreeNode { int val; TreeNode *left; TreeNode *right; Tree…

要求:判断一棵树是否是平衡二叉树 Given a binary tree, determine if it is height-balanced. For . 代码如下: struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x): val(x),left(NULL), right(NULL) {} }; int maxTreeDepth(TreeNode *root) //先求树的深度 { if (NU…

要求:求二叉树的深度(二叉树的深度为最远叶子节点到根节点的距离,即根节点到最远叶子节点的距离) Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. 有两种求解的思路,一种采用DFS的思想,一种采用BFS的思想,如下代码所示: str…

Given a binary tree, return all root-to-leaf paths. Note: A leaf is a node with no children. Example: Input: 1 / \ 2 3 \ 5 Output: ["1->2->5", "1->3"] Explanation: All root-to-leaf paths are: 1->2->5, 1->3 思路为DFS, 只是a…

题目: Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 思路分析: 题意是将二叉树全部左右子数对调,如上图所看到的. 详细做法是,先递归处理左右子树,然后将当前的左右子树对调. 代码实现: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeN…