1、题目
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.
2、思路
题目中让我们判断从根节点到叶子节点是否存在一条路径的和等于```sum```，我们可以认为一棵树是又一个树根和若干子树构成，我们用```sum-root.val```的数值，
再递归判断该根节点的左子树和右子树是否有满足```sum=sum-root.val```的路径.
3、递归实现
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null)
{
return false;
}
// 如果是叶子节点直接返回
if (root.left==null && root.right==null)
{
return root.val == sum;
}
boolean res = true;
// 判断左子树是否满足
if (root.left != null)
{
// 如果是左叶子节点
if (root.left.left == null && root.left.right == null )
{
if ((sum - root.val) == root.left.val)
{
return true;
}
res = false;
}
// 非叶子节点
else{
res = hasPathSum(root.left, sum-root.val);
}
if (res)
{
return res;
}
}
// 如果左子树不满足再判断右子树是否满足
if (root.right != null)
{
// 如果是右叶子节点
if (root.right.left == null && root.right.right == null)
{
if ((sum - root.val) == root.right.val)
{
return true;
}
res = false;
}else
{
res = hasPathSum(root.right, sum-root.val);
}
if (res)
{
return res;
}
}
return res;
}
