LeetCode:Path Sum I &&II

Path Sum

 

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.
第一种方法可以利用栈，非叶子依次入栈，将数值累加起来，一直加到叶子节点此时刚好等于sum，返回true,若不等，处理该叶子节点的父节点，一直到栈空为止。

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {

public boolean hasPathSum(TreeNode root, int sum) {
if(root==null)return false;
if(root.left==null && root.right==null)
{
if(root.val==sum)return true;
else return false;
}
TreeNode p=root,q;
Stack<TreeNode> stack=new Stack<TreeNode>();
stack.add(p);
while(!stack.isEmpty())
{
p=stack.peek();
if(p.left==null && p.right==null)
{
p=stack.pop();
if(stack.isEmpty())return false;
q=stack.peek();
if(q.left==p)q.left=null;
if(q.right==p)q.right=null;
continue;
}
if(p.left!=null)
{
q=p.left;
q.val=q.val+p.val;
if(q.left==null && q.right==null)
{
if(q.val==sum)return true;
else
{
p.left=null;
continue;
}
}
else {
stack.add(q);
continue;
}
}
if(p.right!=null)
{
q=p.right;
q.val=q.val+p.val;
if(q.left==null && q.right==null)
{
if(q.val==sum)return true;
else
{
p.right=null;
continue;
}
}
else {
stack.add(q);
continue;
}

}//else if
}
return false;
}

}

第二种方法直接利用递归，思路简单：

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {

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

}
}

Path Sum II

 

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]
]

第一种方法与I中的方法一类似，需要保存每一步的值进入in list,如果最终值等于sum,将in list 加入最终返回的list中。

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
ArrayList<Integer> copyNewList(ArrayList<Integer> in)
{
int size=in.size();
ArrayList<Integer> out=new ArrayList<Integer>();
Iterator<Integer> it=in.iterator();
while(it.hasNext())out.add(it.next());
out.remove(size-1);
return out;
}

public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
ArrayList<ArrayList<Integer>> list=ne
aa54
w ArrayList<ArrayList<Integer>>();
if(root==null)return list;
Stack<TreeNode> stack=new Stack<TreeNode>();
TreeNode p=root,q;
ArrayList<Integer> in;
if(root.left==null && root.right==null)
{
if(root.val==sum){
in=new ArrayList<Integer>();
in.add(root.val);
list.add(in);
}
return list;
}
stack.add(p);
in=new ArrayList<Integer>();
in.add(p.val);
while(!stack.isEmpty())
{
p=stack.peek();
if(p.left==null && p.right==null)
{
q=stack.pop();
if(stack.isEmpty())return list;
p=stack.peek();
if(p.left==q)p.left=null;
if(p.right==q)p.right=null;
continue;
}
if(p.left!=null)
{
q=p.left;
if(q.left==null && q.right==null)
{
if(p.val+q.val==sum){

in.add(q.val);
list.add(in);
in=copyNewList(in);
}
p.left=null;
continue;
}
else
{
in.add(q.val);
q.val=p.val+q.val;
stack.add(q);
continue;
}
}

if(p.right!=null)
{
q=p.right;
if(q.left==null && q.right==null)
{
if(p.val+q.val==sum){

in.add(q.val);
list.add(in);
in=copyNewList(in);
}
p.right=null;
continue;
}
else {
in.add(q.val);
q.val=p.val+q.val;
stack.add(q);
continue;

}

}// if p.right

}//end while

return list;
}
}

第二种方法类似I中的方法2，利用递归：

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
ArrayList<ArrayList<Integer>> res=new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> in = new ArrayList<Integer>();
public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
if(root == null)return res;
recursion(root, sum);
return res;
}
public void recursion(TreeNode root,int sum)
{
if(root.left == null && root.right == null)
{
in.add(root.val);
int sum1 = 0;
for(int i = 0; i < in.size(); i++)
{
sum1 += in.get(i);
}
if(sum1 == sum)
res.add(new ArrayList<Integer>(in));
return ;
}
in.add(root.val);
if(root.left != null)
{
recursion(root.left,sum);
in.remove(in.size() - 1);
}
if(root.right != null)
{
recursion(root.right,sum);
in.remove(in.size() - 1);
}
}
}