[LeetCode]112. Path Sum&113. Path Sum II

112 . Path Sum

Easy

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.

1ms:

public boolean hasPathSum(TreeNode root, int sum) {
if(root==null)  return false;

return subPath(sum,0,root);
}
private boolean subPath(int sum,int v,TreeNode node){
v += node.val;
if(node.left==null&&node.right==null){
if(v==sum)
return true;
}
if(node.left!=null){
if(subPath(sum,v,node.left))
return true;
}
if(node.right!=null){
if(subPath(sum,v,node.right))
return true;
}
return false;
}

0ms:

public boolean hasPathSum2(TreeNode root, int sum) {
if(root==null) return false;
if(root.left ==null && root.right ==null) return root.val==sum;
return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val);
}

113 . Path Sum II

Medium

Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.

For example:

Given the below binary tree and sum = 22,

5

/ \

4 8

/ / \

11 13 4

/ \ / \

7 2 5 1

return

[

[5,4,11,2],

[5,8,4,5]

]

3ms:

public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(root==null) return result;

List<Integer> link = new LinkedList<Integer>();
subSum(sum,result,link,root);
return result;
}
private void subSum(int v,List<List<Integer>> result,List<Integer> list,TreeNode node){
list.add(node.val);
if(node.left==null&&node.right==null){
if(node.val==v)
result.add(new ArrayList<Integer>(list));
}
if(node.left!=null)
subSum(v-node.val,result,list,node.left);

if(node.right!=null)
subSum(v-node.val,result,list,node.right);

list.remove(list.size()-1);
}

参考：

257. Binary Tree Paths