{"id":10013,"date":"2023-04-29T08:57:33","date_gmt":"2023-04-29T15:57:33","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=10013"},"modified":"2023-04-29T12:14:19","modified_gmt":"2023-04-29T19:14:19","slug":"leetcode-2641-cousins-in-binary-tree-ii","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/uncategorized\/leetcode-2641-cousins-in-binary-tree-ii\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 2641. Cousins in Binary Tree II"},"content":{"rendered":"\n

Given the root<\/code> of a binary tree, replace the value of each node in the tree with the sum of all its cousins’ values<\/strong>.<\/p>\n\n\n\n

Two nodes of a binary tree are cousins<\/strong> if they have the same depth with different parents.<\/p>\n\n\n\n

Return the <\/em>root<\/code> of the modified tree<\/em>.<\/p>\n\n\n\n

Note<\/strong> that the depth of a node is the number of edges in the path from the root node to it.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> root = [5,4,9,1,10,null,7]\nOutput:<\/strong> [0,0,0,7,7,null,11]\nExplanation:<\/strong> The diagram above shows the initial binary tree and the binary tree after changing the value of each node.\n- Node with value 5 does not have any cousins so its sum is 0.\n- Node with value 4 does not have any cousins so its sum is 0.\n- Node with value 9 does not have any cousins so its sum is 0.\n- Node with value 1 has a cousin with value 7 so its sum is 7.\n- Node with value 10 has a cousin with value 7 so its sum is 7.\n- Node with value 7 has cousins with values 1 and 10 so its sum is 11.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> root = [3,1,2]\nOutput:<\/strong> [0,0,0]\nExplanation:<\/strong> The diagram above shows the initial binary tree and the binary tree after changing the value of each node.\n- Node with value 3 does not have any cousins so its sum is 0.\n- Node with value 1 does not have any cousins so its sum is 0.\n- Node with value 2 does not have any cousins so its sum is 0.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • The number of nodes in the tree is in the range [1, 105<\/sup>]<\/code>.<\/li>
  • 1 <= Node.val <= 104<\/sup><\/code><\/li><\/ul>\n\n\n\n
    Solution: Level Order Sum<\/strong><\/h2>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(n)<\/p>\n\n\n\n
    DFS, two passes<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  TreeNode* replaceValueInTree(TreeNode* root) {\n    unordered_map parentSum;\n    parentSum[nullptr] = root->val;\n    vector levelSums;\n    function dfs = [&](TreeNode* node, int depth) {\n      if (!node) return;\n      if (levelSums.size() <= depth) {\n        levelSums.push_back(0);\n      }\n      levelSums[depth] += node->val;\n      if (node->left) {\n        parentSum[node] += node->left->val;\n        dfs(node->left, depth + 1);\n      }\n      if (node->right) {\n        parentSum[node] += node->right->val;\n        dfs(node->right, depth + 1);\n      }\n    };\n    function dfs2 = \n    [&](TreeNode* node, TreeNode* p, int depth) {\n      if (!node) return;\n      node->val = levelSums[depth] - parentSum[p];\n      dfs2(node->left, node, depth + 1);\n      dfs2(node->right, node, depth + 1);\n    };\n    dfs(root, 0);\n    dfs2(root, nullptr, 0);\n    return root;\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    BFS, one+ pass<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  TreeNode* replaceValueInTree(TreeNode* root) {\n    root->val = 0;\n    queue q({root});\n    while (!q.empty()) {\n      vector nodes;\n      int sum = 0;\n      for (int s = q.size(); s; --s) {\n        TreeNode* node = q.front(); q.pop();\n        nodes.push_back(node);\n        if (node->left) {\n           q.push(node->left);\n           sum += node->left->val;\n        }\n        if (node->right) {\n          q.push(node->right);\n          sum += node->right->val;\n        }\n      }\n      for (TreeNode* node : nodes) {\n        int val = sum - (node->left ? node->left->val : 0) \n          - (node->right ? node->right->val : 0);      \n        if (node->left) node->left->val = val;\n        if (node->right) node->right->val = val;\n      }\n    }\n    return root;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
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