Poison

129. Sum Root to Leaf Numbers

发表于
DFS
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class Solution {
    public int sumNumbers(TreeNode root) {
        List<List<Integer>> resultList = new ArrayList<>();
        List<Integer> pathNumList = new ArrayList<>();
        dfs(resultList, pathNumList, root);

        int sum = 0;
        for (List<Integer> numList : resultList) {
            int num = 0;
            for (int n : numList) {
                num = num * 10 + n;
            }
            sum += num;
        }
        return sum;
    }

    private void dfs(List<List<Integer>> resultList, List<Integer> pathNumList, TreeNode root) {
        pathNumList.add(root.val);
        if (root.left == null && root.right == null) {
            // root 为叶子节点，收集
            resultList.add(new ArrayList<>(pathNumList));
        }

        if (root.left != null) {
            dfs(resultList, pathNumList, root.left);
        }
        if (root.right != null) {
            dfs(resultList, pathNumList, root.right);
        }
        pathNumList.remove(pathNumList.size() - 1);
    }
}
DFS
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class Solution {
    public int sumNumbers(TreeNode root) {
        return dfs(root, 0);
    }

    private int dfs(TreeNode root, int preSum) {
        if (root == null) {
            return 0;
        }

        preSum = preSum * 10 + root.val;
        if (root.left == null && root.right == null) {
            return preSum;
        } else {
            return dfs(root.left, preSum) + dfs(root.right, preSum);
        }
    }
}
BFS
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class Solution {
    public int sumNumbers(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        int sum = 0;
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if (node.left == null && node.right == null) {
                sum += node.val;
                continue;
            }

            if (node.left != null) {
                node.left.val = node.val * 10 + node.left.val;
                queue.offer(node.left);
            }
            if (node.right != null) {
                node.right.val = node.val * 10 + node.right.val;
                queue.offer(node.right);
            }
        }

        return sum;
    }
}
Reference

129. Sum Root to Leaf Numbers
剑指 Offer II 049. 从根节点到叶节点的路径数字之和