【Leetcode】Sum Root to Leaf (Sum)
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Given a binary tree, find the maximum path sum.
The path may start and end at any node in the tree.
For example:
Given the below binary tree,
1 / \ 2 3
Return 6
.
递归思路,依次访问每个节点,记录根到叶子的总和,然后全部相加即可。
压栈思路:先左节点全入,和依次递增,然后左出右入。这里只需要用和运算来判断,因为需要找所有路径的总和。
注意:第一次压栈应该压入helper(root, 0), 而不是helper(root, root.val),因为后者会让root.left(root.right)与求和的root.left处于同一个等级,空指针不可避免。而前者是让root.left(root.right)与求和的root处于同一个等级,可以有效避免空指针问题。
public int sumNumbers(TreeNode root) { if(root==null) return 0; return helper(root, 0); } public int helper(TreeNode root, int sum) { if(root==null) return 0; if(root.left==null&&root.right==null) return sum*10+root.val; int left=helper(root.left, sum*10+root.val); int right=helper(root.right, sum*10+root.val); return left+right; }
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