C++实现LeetCode(129.求根到叶节点数字之和)

[leetcode] 129. sum root to leaf numbers 求根到叶节点数字之和

given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

an example is the root-to-leaf path 1->2->3 which represents the number 123.

find the total sum of all root-to-leaf numbers.

note: a leaf is a node with no children.

example:

input: [1,2,3]
1
/ \
2   3
output: 25
explanation:
the root-to-leaf path

1->2

represents the number

12

.
the root-to-leaf path

1->3

represents the number

13

.
therefore, sum = 12 + 13 =

25

.

example 2:

input: [4,9,0,5,1]
4
/ \
9   0
/ \
5   1
output: 1026
explanation:
the root-to-leaf path

4->9->5

represents the number 495.
the root-to-leaf path

4->9->1

represents the number 491.
the root-to-leaf path

4->0

represents the number 40.
therefore, sum = 495 + 491 + 40 =

1026

.

这道求根到叶节点数字之和的题跟之前的求 path sum 很类似，都是利用dfs递归来解，这道题由于不是单纯的把各个节点的数字相加，而是每遇到一个新的子结点的数字，要把父结点的数字扩大10倍之后再相加。如果遍历到叶结点了，就将当前的累加结果sum返回。如果不是，则对其左右子结点分别调用递归函数，将两个结果相加返回即可，参见代码如下：

解法一：

class solution {
public:
    int sumnumbers(treenode* root) {
        return sumnumbersdfs(root, 0);
    }
    int sumnumbersdfs(treenode* root, int sum) {
        if (!root) return 0;
        sum = sum * 10 + root->val;
        if (!root->left && !root->right) return sum;
        return sumnumbersdfs(root->left, sum) + sumnumbersdfs(root->right, sum);
    }
};

我们也可以采用迭代的写法，这里用的是先序遍历的迭代写法，使用栈来辅助遍历，首先将根结点压入栈，然后进行while循环，取出栈顶元素，如果是叶结点，那么将其值加入结果res。如果其右子结点存在，那么其结点值加上当前结点值的10倍，再将右子结点压入栈。同理，若左子结点存在，那么其结点值加上当前结点值的10倍，再将左子结点压入栈，是不是跟之前的 path sum 极其类似呢，参见代码如下：

解法二：

class solution {
public:
    int sumnumbers(treenode* root) {
        if (!root) return 0;
        int res = 0;
        stack<treenode*> st{{root}};
        while (!st.empty()) {
            treenode *t = st.top(); st.pop();
            if (!t->left && !t->right) {
                res += t->val;
            }
            if (t->right) {
                t->right->val += t->val * 10;
                st.push(t->right);
            }
            if (t->left) {
                t->left->val += t->val * 10;
                st.push(t->left);
            }
        }
        return res;
    }
};

github 同步地址：

类似题目：

path sum

binary tree maximum path sum

参考资料：

https://leetcode.com/problems/sum-root-to-leaf-numbers/discuss/41367/non-recursive-preorder-traverse-java-solution

https://leetcode.com/problems/sum-root-to-leaf-numbers/discuss/41452/iterative-c%2b%2b-solution-using-stack-(similar-to-postorder-traversal)

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