path sum

Author: int32bit

README.md

@@ -30,6 +30,7 @@
 + [100 Same Tree](algorithms/SameTree)
 + [101 Symmetric Tree](algorithms/SymmetricTree)
 + [104 Maximum Depth of Binary Tree](algorithms/MaximumDepthofBinaryTree)
++ [112 Path Sum](algorithms/PathSum)
 + [136 Single Number(位运算)](algorithms/SingleNumber)
 + [141 Linked List Cycle](algorithms/LinkedListCycle)
 + [142 Linked List Cycle II](algorithms/LinkedListCycle2)

algorithms/PathSum/README.md

@@ -0,0 +1,38 @@
+## Path Sum
+
+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.
+
+## Solution
+
+递归题
+
+* 如果p是叶子节点，则若`p的值等于sum`，`return true`， 否则`return false`
+* 若p不是叶子节点，则一定存在孩子节点，左孩子或者右孩子满足其中一个和为`sum - p->val`即可,即
+`return hasPath(p->left, sum - p->val) || hasPath(p->right, sum - p->right)`
+
+## Code
+```c
+bool hasPathSum(struct TreeNode *root, int sum) {
+	if (root == NULL)
+		return false;
+	if (root->left == NULL && root->right == NULL)
+		return root->val == sum;
+	if (!root->left || !hasPathSum(root->left, sum - root->val)) {
+		return hasPathSum(root->right, sum - root->val);
+	}
+	return true;
+}
+```

algorithms/PathSum/solve.c

@@ -0,0 +1,51 @@
+#include <stdio.h>
+#include <stdbool.h>
+#include <stdlib.h>
+
+struct TreeNode {
+	int val;
+	struct TreeNode *left;
+	struct TreeNode *right;
+};
+bool hasPathSum(struct TreeNode *root, int sum) {
+	if (root == NULL)
+		return false;
+	if (root->left == NULL && root->right == NULL)
+		return root->val == sum;
+	if (!root->left || !hasPathSum(root->left, sum - root->val)) {
+		return hasPathSum(root->right, sum - root->val);
+	}
+	return true;
+}
+
+struct TreeNode *mk_node(int val)
+{
+	struct TreeNode *p = malloc(sizeof(*p));
+	p->val = val;
+	p->left = p->right = NULL;
+	return p;
+}
+void mk_child(struct TreeNode *root, struct TreeNode *left, struct TreeNode *right)
+{
+	root->left = left;
+	root->right = right;
+}
+int main(int argc, char **argv)
+{
+	struct TreeNode *root = mk_node(5);
+	mk_child(root, mk_node(4), mk_node(8));
+	struct TreeNode *left = root->left;
+	struct TreeNode *right = root->right;
+	mk_child(left, mk_node(11), NULL);
+	mk_child(left->left, mk_node(7), mk_node(2));
+	mk_child(right, mk_node(13), mk_node(4));
+	mk_child(right->right, NULL, mk_node(1));
+	printf("%d\n", hasPathSum(root, 22));
+	printf("%d\n", hasPathSum(root, 26));
+	printf("%d\n", hasPathSum(root, 18));
+	printf("%d\n", hasPathSum(root, 27));
+	printf("%d\n", hasPathSum(root, 37));
+	printf("%d\n", hasPathSum(root, 100));
+	printf("%d\n", hasPathSum(root, 0));
+	return 0;
+}