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.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
//递归实现
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) {
return false;
}
if (root.left == null && root.right == null) {
return sum == root.val;
}
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}
//非递归:DFS 使用栈
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) {
return false;
}
Stack<TreeNode> stack = new Stack<TreeNode>();
Stack<Integer> sums = new Stack<Integer>();
stack.push(root);
sums.push(sum);
while(!stack.isEmpty()) {
TreeNode node = stack.pop();
int value = sums.pop();
if (node.left == null && node.right == null && node.val == value) {
return true;
}
if (node.left != null) {
stack.push(node.left);
sums.push(value - node.val);
}
if (node.right != null){
stack.push(node.right);
sums.push(value - node.val);
}
}
return false;
}
}