Binary Tree Path Sum 解题报告

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Binary Tree Path Sum

Description

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]
]

实现思路

进行先序遍历，遍历到每个叶子节点的时候，判断sum是否等于target，相等则添加到结果集中。
注意在回溯的时候，要删除当前层次添加的节点。

/**
 * 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
     */
    int target;
    List<List<Integer>> ret = new ArrayList<>();
    public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
        if(root == null){
            return ret; 
        }
        this.target = target; 
        List<Integer> path = new ArrayList<>();
        helper(path, root,0);
        return ret;
    }

    public void helper(List<Integer> path, TreeNode root, int sum){
        sum += root.val;
        path.add(root.val);
        if(root.left == null && root.right == null && sum == target){
            List<Integer> newList = new ArrayList<>();
            newList.addAll(path);
            ret.add(newList);
        }
        if(root.left != null){
            helper(path,root.left,sum);
        }
        if(root.right != null){
            helper(path,root.right,sum);
        }
        path.remove(path.size() -1);
    }    

}