Given a binary tree, find the leftmost value in the last row of the tree.
Example 1:
Input:
2
/ \
1 3
Output:
1
Example 2:
Input:
1
/ \
2 3
/ / \
4 5 6
/
7
Output:
7
Note: You may assume the tree (i.e., the given root node) is not NULL.
题目链接:https://leetcode.com/problems/find-bottom-left-tree-value/description/
题目分析:裸BFS,先右后左
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int findBottomLeftValue(TreeNode root) {
Queue<TreeNode> q = new LinkedList<>();
q.add(root);
int ans = root.val;
while (!q.isEmpty()) {
TreeNode cur = q.poll();
ans = cur.val;
if (cur.right != null) {
q.add(cur.right);
}
if (cur.left != null) {
q.add(cur.left);
}
}
return ans;
}
}

本文介绍了一种解决LeetCode上特定问题的方法:寻找二叉树最底层最左边的节点值。通过使用从右到左的广度优先搜索(BFS),可以有效地找到目标节点。
943

被折叠的 条评论
为什么被折叠?



