Problem Link: http://oj.leetcode.com/problems/binary-tree-preorder-traversal/ Even iterative solution is easy, just use a stack storing the nodes not visited. Each iteration, pop a node and visited it, then push its right child and then left child in…

Problem Link: http://oj.leetcode.com/problems/binary-tree-postorder-traversal/ The post-order-traversal of a binary tree is a classic problem, the recursive way to solve it is really straightforward, the pseudo-code is as follows. RECURSIVE-POST-ORDE…

Problem Link: https://oj.leetcode.com/problems/binary-tree-level-order-traversal/ Traverse the tree level by level using BFS method. # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # se…

Problem Link: https://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ Just BFS from the root and for each level insert a list of values into the result. # Definition for a binary tree node # class TreeNode: # def __init__(self, x):…

Problem Link: https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/ Use BFS from the tree root to traverse the tree level by level. The python code is as follows. # Definition for a binary tree node # class TreeNode: # def __init__(s…

Problem Link: http://oj.leetcode.com/problems/binary-tree-maximum-path-sum/ For any path P in a binary tree, there must exists a node N in P such that N is the ancestor node of all other nodes in P. We call such N as the root of P, or P roots at N. T…

Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? Subscribe to see which companies asked this…

Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3} 1 \ 2 / 3 return [1,2,3]. 前序遍历二叉树,只不过题目的要求是尽量不要使用递归,当然我还是先用递归写了一个: /** * Definition for a binary tree node. * struct TreeNode { * int val…

Binary Tree Preorder Traversal Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? Solution: 递归…

Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. 二叉树的先序遍历.递归算法太简单了,重点讨论一下非递归的. 我自己写的时候一直纠结于何时弹出结点.后来看了几个版本的,发现可以跳过这个部分. 贴上我最喜欢的版本,逻辑最清楚 //我最喜欢的版本 逻辑清晰 vector<i…