内容简介:Given a binary tree, find the leftmost value in the last row of the tree.Example 1:Input:
Given a binary tree, find the leftmost value in the last row of the tree.
Example 1:
Input:
/ \
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.
难度:medium
题目:
给定二叉树,找出其最左边的结点。
注意:
你可以认为树一定不为NULL
思路:
BFS, 层次遍历(先入右子树,再入左子树)。
Runtime: 4 ms, faster than 71.59% of Java online submissions for Find Bottom Left Tree Value.
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public int findBottomLeftValue(TreeNode root) { Queue<TreeNode> queue = new LinkedList<>(); queue.add(root); TreeNode node = root; while (!queue.isEmpty()) { node = queue.poll(); if (node.right != null) { queue.add(node.right); } if (node.left != null) { queue.add(node.left); } } return node.val; } }
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