*Path Sum
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.
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.
解法一:递归的解法(reference:http://www.cnblogs.com/springfor/p/3879825.html)
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 }
解法二:非递归的解法(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 }