标签：private travel sum == java des parent war tco
437. Path Sum III
You are given a binary tree in which each node contains an integer value.
Find the number of paths that sum to a given value.
The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
10
/ 5 -3
/ \ 3 2 11
/ \ 3 -2 1
Return 3. The paths that sum to 8 are:
1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11
package leetcode.easy;
/**
* Definition for a binary tree node. public class TreeNode { int val; TreeNode
* left; TreeNode right; TreeNode(int x) { val = x; } }
*/
public class PathSumIII {
int count = 0;
private void helper(TreeNode root, int sum) {
if (root == null) {
return;
}
if (root.val == sum) {
count++;
}
if (root.left != null) {
helper(root.left, sum - root.val);
}
if (root.right != null) {
helper(root.right, sum - root.val);
}
}
public int pathSum(TreeNode root, int sum) {
if (root == null) {
return count;
}
helper(root, sum);
if (root.left != null) {
pathSum(root.left, sum);
}
if (root.right != null) {
pathSum(root.right, sum);
}
return count;
}
@org.junit.Test
public void test() {
int sum = 8;
TreeNode tn11 = new TreeNode(10);
TreeNode tn21 = new TreeNode(5);
TreeNode tn22 = new TreeNode(-3);
TreeNode tn31 = new TreeNode(3);
TreeNode tn32 = new TreeNode(2);
TreeNode tn34 = new TreeNode(11);
TreeNode tn41 = new TreeNode(3);
TreeNode tn42 = new TreeNode(-2);
TreeNode tn44 = new TreeNode(1);
tn11.left = tn21;
tn11.right = tn22;
tn21.left = tn31;
tn21.right = tn32;
tn22.left = null;
tn22.right = tn34;
tn31.left = tn41;
tn31.right = tn42;
tn32.left = null;
tn32.right = tn44;
tn34.left = null;
tn34.right = null;
tn41.left = null;
tn41.right = null;
tn42.left = null;
tn42.right = null;
tn44.left = null;
tn44.right = null;
System.out.println(pathSum(tn11, sum));
}
}
标签：private travel sum == java des parent war tco
原文地址：https://www.cnblogs.com/denggelin/p/11948833.html
