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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.
分治法可以解决，传入下层的sum为原来的sum-root.val即可
1 public boolean hasPathSum(TreeNode root, int sum) { 2 if(root==null) 3 return false; 4 if(root.left==null && root.right==null) 5 return root.val==sum; 6 return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val); 7 }
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]
]
II就是要返回所有可能的path. 可以用分治法的思想去实现（把根节点加到左子树得到的list和右子树得到的list的第一位），不过较慢，因为要结果返回给上层。用单纯的dfs回溯也能很好地实现，而且较快。
分治法：
1 public List<List<Integer>> pathSum(TreeNode root, int sum) { 2 List<List<Integer>> re = new ArrayList<List<Integer>>(); 3 if(root==null) 4 return re; 5 if(root.left==null && root.right==null && root.val==sum) { 6 List<Integer> temp = new ArrayList<Integer>(); 7 temp.add(root.val); 8 re.add(temp); 9 return re; 10 } 11 List<List<Integer>> left = pathSum(root.left, sum-root.val); 12 List<List<Integer>> right = pathSum(root.right, sum-root.val); 13 if(left.size()>0) 14 for(int i=0;i<left.size();i++) { 15 left.get(i).add(0,root.val); 16 re.add(left.get(i)); 17 } 18 if(right.size()>0) 19 for(int i=0;i<right.size();i++) { 20 right.get(i).add(0,root.val); 21 re.add(right.get(i)); 22 } 23 return re; 24 }
dfs：
public List<List<Integer>> pathSum(TreeNode root, int sum) { List<List<Integer>> re = new ArrayList<List<Integer>>(); if(root==null) return re; List<Integer> path = new ArrayList<Integer>(); collect(re, path, root, sum); return re; } public void collect(List<List<Integer>> re, List<Integer> path, TreeNode rt, int v) { if(rt.left==null && rt.right==null && rt.val==v) { List<Integer> temp = new ArrayList<Integer>(path); temp.add(rt.val); re.add(temp); return; } path.add(rt.val); if(rt.left!=null) collect(re,path,rt.left,v-rt.val); if(rt.right!=null) collect(re, path, rt.right, v-rt.val); path.remove(path.size()-1); }
[Leetcode][JAVA] Path Sum I && II
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原文地址：http://www.cnblogs.com/splash/p/4397871.html
