112. Path Sum二叉樹路徑和
阿新 • • 發佈:2018-03-11
aps exist display splay term post ase urn 數據結構
[抄題]:
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and sum = 22
,
5 / 4 8 / / 11 13 4 / \ 7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2
which sum is 22.
[暴力解法]:
時間分析:
空間分析:
[奇葩輸出條件]:
字符串中間還要加符號,頭一次見
[奇葩corner case]:
兩端都是空,只有root一個點也是能求和的,第一次見。下次二叉樹求和的時候只要有點就能加
[思維問題]:
不知道怎麽利用sum:變成參數即可
[一句話思路]:
還是沒有匯總的traverse,把sum參與進來,把sum參數化變成一個參數就行了
[輸入量]:空: 正常情況:特大:特小:程序裏處理到的特殊情況:異常情況(不合法不合理的輸入):
[畫圖]:
[一刷]:
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分鐘肉眼debug的結果]:
[總結]:
[復雜度]:Time complexity: O(n) Space complexity: O(n)
[英文數據結構或算法,為什麽不用別的數據結構或算法]:
[關鍵模板化代碼]:
[其他解法]:
[Follow Up]:
[LC給出的題目變變變]:
path sum後續系列
[代碼風格] :
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * }View Code*/ class Solution { public boolean hasPathSum(TreeNode root, int sum) { //root is null if (root == null) { return false; } //only root is not null if (root.left == null && root.right == null) { return root.val == sum; } //remaning sum in left or sum in right return (hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val)); } }
112. Path Sum二叉樹路徑和