lintcode376- Binary Tree Path Sum- easy

Given a binary tree, find all paths that sum of the nodes in the path equals to a given number target.

A valid path is from root node to any of the leaf nodes.

Example

Given a binary tree, and target = 5:

     1
    / \
   2   4
  / \
 2   3

return

[
  [1, 2, 2],
  [1, 4]
]

 

分治法。拆解为在子树里找凑出(target-root.value)的值的问题，一直推到叶子节点来解决。如果叶子节点的val正好target重合了，就说明root到这个叶子节点这一路是可行的。题目还关注了下List数据结构的操作，逻辑理清楚。

 

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */


public class Solution {
    /*
     * @param root: the root of binary tree
     * @param target: An integer
     * @return: all valid paths
     */
    public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
        // write your code here
        
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        
        if (root == null) {
            return result;
        }
        
        if (root.left == null && root.right == null && root.val == target) {
            List<Integer> newPath = new ArrayList<Integer>();
            newPath.add(root.val);
            result.add(newPath);
            return result;
        }
        
        List<List<Integer>> left = binaryTreePathSum(root.left, target - root.val);
        List<List<Integer>> right = binaryTreePathSum(root.right, target - root.val);
        
        for (List<Integer> path : left) {
            List<Integer> newPath = new ArrayList<Integer>();
            newPath.add(root.val);
            newPath.addAll(path);
            result.add(newPath);
        }
        
        for (List<Integer> path : right) {
            List<Integer> newPath = new ArrayList<Integer>();
            newPath.add(root.val);
            newPath.addAll(path);
            result.add(newPath);
        }
        
        return result;
    }
    
}