第二道树的题目,依旧不会做,谷歌经验. 题目解释: give you a tree , judge if it is a symmetric tree. 思路:我以为要写个中序遍历(进阶学习非递归算法)什么的,Wrong Answer. 解题思路:如果 左子树的值 等于 右子树的值,并且该左子树的左子树 等于 该右子树的右子树,并且该左子树的右子树 等于 该右子树的左子树 时,该树为symmetric.(如果tree == null || tree中只有一个节点,return true) 代码:…

LeetCode:Binary Tree Level Order Traversal Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level ord…

LeetCode: Binary Tree Traversal 题目:树的先序和后序. 后序地址:https://oj.leetcode.com/problems/binary-tree-postorder-traversal/ 先序地址:https://oj.leetcode.com/problems/binary-tree-preorder-traversal/ 后序算法:利用栈的非递归算法.初始时,先从根节点一直往左走到底,并把相应的元素进栈:在循环里每次都取出栈顶元素,如果该栈顶元素的右…

必须承认,一開始这道题我是不会做的.由于我心目中的树遍历仅仅能用一个节点发起.多么天真而无知. 我想不通如何同一时候遍历两颗子树.由于根节点一定是一个啊.但是,作为对称轴上的它.从一開始就不应该被考虑,他的左右孩子,不是非常自然的形成了两个遍历的入口吗?可见无知是多么的可怕. bool helper(TreeNode *left, TreeNode *right){ if(left == NULL && right == NULL) return true; if(left == NULL…

Given a binary tree, return the vertical order traversal of its nodes' values. (ie, from top to bottom, column by column). If two nodes are in the same row and column, the order should be from left to right. Examples: Given binary tree [3,9,20,null,n…

Given a binary tree, find the length of the longest consecutive sequence path. The path refers to any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The longest consecutive path need to be from p…

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"] 这道题给我们一个二叉树,让我们返回所有根到叶节点的路径,跟之前那道Path Sum II 二叉树路径之和之二很类似,比那道稍微简单一…

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. For example:Given the following binary tree, 1 <--- / \ 2 3 <--- \ \ 5 4 <--- You should return [1, 3,…

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return…

Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could you do it iteratively? 经典题目,求二叉树的后序遍历的非递归方法,跟前序,中序,层序一样都需要用到栈,后续的…