題目:
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.
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.
題解:
還是對樹的操作,遞迴的解法:
1 public boolean hasPathSum(TreeNode root, int sum) {
2 if(root == null)
3 return false;
4
5 sum -= root.val;
6 if(root.left == null && root.right==null)
7 return sum == 0;
8 else
9 return hasPathSum(root.left,sum) || hasPathSum(root.right,sum);
10 }
2 if(root == null)
3 return false;
4
5 sum -= root.val;
6 if(root.left == null && root.right==null)
7 return sum == 0;
8 else
9 return hasPathSum(root.left,sum) || hasPathSum(root.right,sum);
10 }
非遞迴的解法(Reference:http://www.programcreek.com/2013/01/leetcode-path-sum/):
1 public boolean hasPathSum(TreeNode root, int sum) {
2 if(root == null) return false;
3
4 LinkedList<TreeNode> nodes = new LinkedList<TreeNode>();
5 LinkedList<Integer> values = new LinkedList<Integer>();
6
7 nodes.add(root);
8 values.add(root.val);
9
10 while(!nodes.isEmpty()){
11 TreeNode curr = nodes.poll();
12 int sumValue = values.poll();
13
14 if(curr.left == null && curr.right == null && sumValue==sum){
15 return true;
16 }
17
18 if(curr.left != null){
19 nodes.add(curr.left);
20 values.add(sumValue+curr.left.val);
21 }
22
23 if(curr.right != null){
24 nodes.add(curr.right);
25 values.add(sumValue+curr.right.val);
26 }
27 }
28
29 return false;
30 }
2 if(root == null) return false;
3
4 LinkedList<TreeNode> nodes = new LinkedList<TreeNode>();
5 LinkedList<Integer> values = new LinkedList<Integer>();
6
7 nodes.add(root);
8 values.add(root.val);
9
10 while(!nodes.isEmpty()){
11 TreeNode curr = nodes.poll();
12 int sumValue = values.poll();
13
14 if(curr.left == null && curr.right == null && sumValue==sum){
15 return true;
16 }
17
18 if(curr.left != null){
19 nodes.add(curr.left);
20 values.add(sumValue+curr.left.val);
21 }
22
23 if(curr.right != null){
24 nodes.add(curr.right);
25 values.add(sumValue+curr.right.val);
26 }
27 }
28
29 return false;
30 }