Tree with Pure Recursion and Backtracking Summary
Pure Recursion: 该点作为root,从后面问题继承的,如何给回之前以至于找到解
curResult.left(), after curPoint
curResult.right(), after curPoint
小心你的base case, cur == null , AND, cur.left == null & cur.right == null
小心你的cur.left or cur.right 小心cur有没有,cur.left & cur.right 有么有
Backtracking: 该点作为ending point, 如何利用之前继承的,往下找到解
f(curPoint, resultInforBeforeCurPoint, curResultInforBeforeCurPoint)
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