//Definition for a binary tree node.structTreeNode {intval; TreeNode*left; TreeNode*right; TreeNode(intx) : val(x), left(NULL), right(NULL) {} }; 2.遍历 a.递归先序: //递归先序: 中左右。PS:中序-左中右,后序-左右中,调换cout的位置即可voidNL
*/classSolution{public:int ret=0;voiddfs(TreeNode*root,int sum){if(root->left==nullptr&&root->right==nullptr){ret+=sum;return;}if(root->left)dfs(root->left,sum*10+root->left->val);if(root->right)dfs(root->right,sum*10+root->right->val);}intsumNumbers(TreeNode*root){dfs(root...
// 二叉链 struct BinaryTreeNode { struct BinTreeNode* _pLeft; // 指向当前节点左孩子 struct BinTreeNode* _pRight; // 指向当前节点右孩子 BTDataType _data; // 当前节点值域 } // 三叉链 struct BinaryTreeNode { struct BinTreeNode* _pParent; // 指向当前节点的双亲 struct BinTreeNode* _p...
a_node = BinaryTree('a') a_node.insert_left('b') a_node.insert_right('c') b_node = a_node.left_child b_node.insert_right('d') d_node = b_node.right_child c_node = a_node.right_child c_node.insert_left('e') c_node.insert_right('f') e_node = c_node.left_child f...
# 二叉树节点定义classTreeNode:def__init__(self,val):self.val=val self.left=None self.right=None # 二叉树的DFS遍历 defdfs_binary_tree(root):ifroot is None:returnprint(root.val,end=' ')dfs_binary_tree(root.left)dfs_binary_tree(root.right)# 构造二叉树 root=TreeNode(1)root.left=Tree...
b) Print the popped item, set current = popped_item->right c) Go to step 3. 5) If current is NULL and stack is empty then we are done. 代码实现: // C++ program to print inorder traversal // using stack. #include<bits/stdc++.h> using namespace std; /* A binary tree Node ...
c语言刷 DFS题记录 144. 二叉树的前序遍历 /** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * };*//** * Note: The returned array must be malloced, assume caller calls free().*///递归voidPreorder(...
reConstructBinaryTree(pre[root + 1: ], tin[root + 1: ]) return res 二叉树的子结构 题目描述:输入两棵二叉树A,B,判断B是不是A的子结构。(ps:我们约定空树不是任意一个树的子结构)如图,第二棵树为第一棵树的子树。 思考:我们是要判断B是不是A的子结构,则B是子树,A为原树。我们可以先找到B的...
层次遍历从低往root结点输出,如Given binary tree{3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [15,7], [9,20], [3] ] 最简单方法通过层次遍历BFS调用队列后逆序倒置vector容器即可。 /**
3.Maximum Depth of Binary Tree - 求二叉树的深度 DFS 4.Balanced Binary Tree - 判断平衡二叉树 DFS 5.Path Sum - 二叉树路径求和判断DFS 题目概述: Given a binary tree, return thelevel ordertraversal of its nodes' values. (ie, from left to right, level by level). ...