Path Sum II

https://leetcode.com/problems/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.

Thoughts

DFS整棵树.

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private void dfs(TreeNode node, int residual, List<Integer> path, List<List<Integer>> res) {
        if (node == null) {
            return;
        }

        path.add(node.val);
        residual -= node.val;
        if (node.left == null && node.right == null && residual == 0) {
            res.add(new ArrayList<>(path));
        } 

        dfs(node.left, residual, path, res);
        dfs(node.right, residual, path, res);

        residual += node.val;
        path.remove(path.size() - 1);
    }

    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<>();
        dfs(root, sum, new ArrayList<>(), res);
        return res;
    }
}

Analysis

Errors:

1. 当是叶子节点且应该加入res忘了把最后一个结点加入path.

每个结点都会被访问，时间复杂度O(n), n为结点数。

Ver. 2

https://discuss.leetcode.com/topic/31698/java-solution-iterative-and-recursive

看着和inorder很像, 区别在于当退回到中间节点时, 不能直接pop, 否则path将不再包含它, 而是要检查它的右子树是否存在, 并把右子树pushLeft. 但如果每次都pushLeft则无法正常让中间节点退出, 因此还要把上个退出的节点记下来, 看是否和中间节点的right一致, 当一致时意味着右子树已经查过了, 不需要再pushLeft了.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        stack<TreeNode*> s;
        vector<vector<int>> r;
        vector<int> p;
        int psum = 0;
        const auto push_left = [&](TreeNode *cur) {
            while (cur != NULL) {
                psum += cur->val;
                p.push_back(cur->val);
                s.push(cur);
                cur = cur->left;
            }
        };
        TreeNode *pre;
        push_left(root);
        while (!s.empty()) {
            const auto cur = s.top();
            if (cur->right != NULL && cur->right != pre) {
                push_left(cur->right);
                continue;
            }
            if (cur->left == NULL && cur->right == NULL && psum == sum) {
                r.push_back(p);
            }
            s.pop();
            psum -= cur->val;
            p.pop_back();
            pre = cur;
        }
        return r;
    }
};