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.
这个题目其实就是二叉树的遍历,我采用深度遍历
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void depthFirstSearch(TreeNode* root, int sum, bool &tag) { if (root==NULL)return; if (root->left) { depthFirstSearch(root->left, sum - root->val, tag); depthFirstSearch(root->right, sum - root->val, tag); } else { if (root->right) { if (root->right->right == NULL&&root->right->left == NULL) { if (sum - root->val - root->right->val == 0)tag = true; return; } else { depthFirstSearch(root->right->left, sum - root->val - root->right->val, tag); depthFirstSearch(root->right->right, sum - root->val - root->right->val, tag); } } else { if (0 == sum-root->val)tag = true; } } } bool hasPathSum(TreeNode* root, int sum) { bool tag = false; depthFirstSearch(root, sum, tag); return tag; } };