这个题的思路其实跟[LeetCode] 199. Binary Tree Right Side View_ Medium tag: BFS, Amazon里面我提到的left side view一样的思路, 只是返回的时候返回最后一个元素即可. 1. Constraints 1) root cannot be None, 所以edge case就是 1 2, Ideas BFS: T: O(n), S: O(n) n is the number of the...
* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public int sumOfLeftLeaves(TreeNode root) { int sum=0; //如果为null直接返回0 if(root==null){ return 0;...
Note: You may assume the tree (i.e., the given root node) is not NULL. 分析:广搜,层序遍历,保存每层的第一个值,最后的即为所求值。 1/**2* Definition for a binary tree node.3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x)...
Given a binary tree, find the leftmost value in the last row of the tree. 就是拿到最后一层的第一个元素。 这个元素是最左下角的元素,不是最左侧的元素。 如果想实现 其实也很简单 就是维护更新每一层的第一个元素。 class Solution { public int findBottomLeftValue(TreeNode root) { if (root =...
LeetCode 404 sum of left leaves Find the sum of all left leaves in a given binary tree think about this, when we reached leaf, how are we gonna to know if it is left leaf or not? of course we can modify the iterate version of preorder traverse, it’s very simple....