标签:scribe and cat node font style turn break dtree
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
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解答
/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ struct TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize) { struct TreeNode *node; int i; if(0 >= preorderSize||0 >= inorderSize){ return NULL; } node = (struct TreeNode*)malloc(sizeof(struct TreeNode)); node->val = preorder[0]; for(i = 0; i < inorderSize; i++){ if(inorder[i] == preorder[0]){ break; } } node->left = buildTree(preorder + 1, i, inorder, i); node->right = buildTree(preorder + i + 1, preorderSize - 1 - i, inorder + i + 1, preorderSize - 1 - i); return node; }
LeetCode OJ 105. Construct Binary Tree from Preorder and Inorder Traversal
标签:scribe and cat node font style turn break dtree
原文地址:http://www.cnblogs.com/YuNanlong/p/6087540.html