标签：junit int pac bool new leetcode code sum OLE
112. 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.
Note: A leaf is a node with no children.
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.
package leetcode.easy;
public class PathSum {
@org.junit.Test
public void test() {
int sum = 22;
TreeNode tn11 = new TreeNode(5);
TreeNode tn21 = new TreeNode(4);
TreeNode tn22 = new TreeNode(8);
TreeNode tn31 = new TreeNode(11);
TreeNode tn33 = new TreeNode(13);
TreeNode tn34 = new TreeNode(4);
TreeNode tn41 = new TreeNode(7);
TreeNode tn42 = new TreeNode(2);
TreeNode tn46 = new TreeNode(1);
tn11.left = tn21;
tn11.right = tn22;
tn21.left = tn31;
tn21.right = null;
tn22.left = tn33;
tn22.right = tn34;
tn31.left = tn41;
tn31.right = tn42;
tn33.left = null;
tn33.right = null;
tn34.left = null;
tn34.right = tn46;
tn41.left = null;
tn41.right = null;
tn42.left = null;
tn42.right = null;
tn46.left = null;
tn46.right = null;
System.out.println(hasPathSum(tn11, sum));
}
public boolean hasPathSum(TreeNode root, int sum) {
if (null == root) {
return false;
} else if (null == root.left && null == root.right && 0 == sum - root.val) {
return true;
} else {
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}
}
标签：junit int pac bool new leetcode code sum OLE
原文地址：https://www.cnblogs.com/denggelin/p/11619789.html
