Path Sum I

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.

递归法

复杂度

时间 O(b^(h+1)-1) 空间 O(h) 递归栈空间 对于二叉树b=2

思路

要求是否存在一个累加为目标和的路径，我们可以把目标和减去每个路径上节点的值，来进行递归。

代码

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

深度优先搜索

复杂度

时间 O(b^(h+1)-1) 空间 O(h) 递归栈空间 对于二叉树b=2

思路

基本的深度优先搜索，思路和上题一样用目标和减去路径上节点的值，不过要记录下搜索时的路径，把这个临时路径代入到递归里。

代码

public class Solution {
    
    List<List<Integer>> res;
    
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        res = new LinkedList<List<Integer>>();
        List<Integer> tmp = new LinkedList<Integer>();
        if(root!=null) helper(root, tmp, sum);
        return res;
    }
    
    private void helper(TreeNode root, List<Integer> tmp, int sum){
        if(root.val == sum && root.left==null && root.right==null){
            tmp.add(root.val);
            List<Integer> one = new LinkedList<Integer>(tmp);
            res.add(one);
            tmp.remove(tmp.size()-1);
        } else {
            tmp.add(root.val);
            if(root.left!=null) helper(root.left, tmp, sum - root.val);
            if(root.right!=null) helper(root.right, tmp, sum - root.val);
            tmp.remove(tmp.size()-1);
        }
    } 
}