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...
1/**2* Definition for a binary tree node.3* public class TreeNode {4* int val;5* TreeNode left;6* TreeNode right;7* TreeNode() {}8* TreeNode(int val) { this.val = val; }9* TreeNode(int val, TreeNode left, TreeNode right) {10* this.val = val;11* this.left = left;1...
使用层序遍历思想 3、代码 1intfindBottomLeftValue(TreeNode*root) {2if(root ==NULL)3return0;4queue<TreeNode*>q;5q.push(root);67intval =0;8while(!q.empty()) {9intsize =q.size();10for(inti =0; i < size; i++) {11TreeNode *node =q.front();12if(node->left !=NULL)13q.pu...
https://leetcode.com/problems/find-bottom-left-tree-value/ 用BFS,层次遍历 /*** Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * }*/publicclassSolution {publicintfindBottomLeftValue(Tree...