[抄题]:
Find the sum of all left leaves in a given binary tree.
Example:
3 / 9 20 / 15 7 There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.
[暴力解法]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
root.left root.right只是递归的过程,必须要真正到了叶子节点才能求和
[一句话思路]:
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- 左子树第一步先检验是否到了叶子节点,类似于特殊判断的思想,涨经验
- 在全树求和的题要+= 才行,不是递归一个算式就完了
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
root.left root.right只是递归的过程,必须要真正到了叶子节点才能求和
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
/** * 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) { //corner case if (root == null) { return 0; } int sum = 0; //left if (root.left != null) { if (root.left.left == null && root.left.right == null) { sum += root.left.val; }else { sum += sumOfLeftLeaves(root.left); } } //right sum += sumOfLeftLeaves(root.right); //return return sum; } }