LeetCode545.Boundary-of-Binary-Tree】的更多相关文章

Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes. Left boundary is defined as the path from root to the…
[抄题]: Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes. Left boundary is defined as the path from root…
Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes.  Left boundary is defined as the path from root to th…
Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes. Left boundary is defined as the path from root to the…
原题链接在这里:https://leetcode.com/problems/find-leaves-of-binary-tree/#/description 题目: 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…
二叉树基础 满足这样性质的树称为二叉树:空树或节点最多有两个子树,称为左子树.右子树, 左右子树节点同样最多有两个子树. 二叉树是递归定义的,因而常用递归/DFS的思想处理二叉树相关问题,例如LeetCode题目 104. Maximum Depth of Binary Tree: // 104. Maximum Depth of Binary Tree int maxDepth(TreeNode* root) { ; +max(maxDepth(root->left),maxDepth(roo…
Given a binary tree, find all leaves and then remove those leaves. Then repeat the previous steps until the tree is empty. Example:Given binary tree 1 / \ 2 3 / \ 4 5 Returns [4, 5, 3], [2], [1]. Explanation: 1. Remove the leaves [4, 5, 3] from the t…
Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes. Left boundary is defined as the path from root to the…
One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as #. _9_ / \ 3 2 / \ / \ 4 1 # 6 / \ / \ / \ # # # # # # For…
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…