JavaScript算法实例之求二叉树从根到叶的所有路径和

        5
      /   \
     3     8
    / \   / \
   1   4 7   9

5 -> 3 -> 1  为 531
5 -> 3 -> 4  为 534
5 -> 8 -> 7  为 587
5 -> 8 -> 9  为 589

class TreeNode {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

function sumPaths(node, currentSum = 0) {
  if (!node) return 0;

  currentSum = currentSum * 10 + node.value;

  // 如果是叶子节点，直接返回当前和
  if (!node.left && !node.right) return currentSum;

  // 否则返回左右子树的和
  return sumPaths(node.left, currentSum) + sumPaths(node.right, currentSum);
}

// 使用上面的树作为示例
const root = new TreeNode(5);
root.left = new TreeNode(3);
root.right = new TreeNode(8);
root.left.left = new TreeNode(1);
root.left.right = new TreeNode(4);
root.right.left = new TreeNode(7);
root.right.right = new TreeNode(9);

console.log(sumPaths(root)); // 输出: 2241

function dfs(node) {
  if (node == null) return;

  console.log(node.value);  // 处理当前节点
  dfs(node.left);  // 递归访问左子节点
  dfs(node.right);  // 递归访问右子节点
}

function dfs(graph, start) {
  let stack = [];
  let visited = new Set();

  stack.push(start);

  while (stack.length) {
    let node = stack.pop();
    if (!visited.has(node)) {
      console.log(node);  // 处理当前节点
      visited.add(node);

      // 添加未访问的邻接节点到堆栈
      for (let neighbor of graph[node]) {
        if (!visited.has(neighbor)) {
          stack.push(neighbor);
        }
      }
    }
  }
}