Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.
從前序以及中序的結果中構造二叉樹，這里保證了不會有兩個相同的數字，用遞歸構造就比較方便了：

1 class Solution { 2 public: 3 TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { 4 if(!preorder.size()) return NULL; 5 return createTree(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);//邊界條件應該想清楚 6 } 7 8 TreeNode* createTree(vector<int>& preOrder, int preBegin, int preEnd, vector<int>& inOrder, int inBegin, int inEnd) 9 { 10 if(preBegin > preEnd) return NULL; 11 int rootVal = preOrder[preBegin]; 12 int mid; 13 for(int i = inBegin; i <= inEnd; ++i) 14 if(inOrder[i] == rootVal){ 15 mid = i; 16 break; 17 } 18 int len = mid - inBegin; //左邊區間的長度為mid - inBegin; 19 TreeNode * left = createTree(preOrder, preBegin + 1, preBegin + len, inOrder, inBegin, mid - 1); 20 TreeNode * right = createTree(preOrder, preBegin + len + 1, preEnd, inOrder, mid + 1, inEnd); 21 TreeNode * root = new TreeNode(rootVal); 22 root->left = left; 23 root->right = right; 24 return root; 25 } 26 };

?java版本的代碼如下所示，方法上與上面的沒什么區別：

1 public class Solution { 2 public TreeNode buildTree(int[] preorder, int[] inorder) { 3 return createTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1); 4 } 5 6 public TreeNode createTree(int[] preorder, int preBegin, int preEnd, int[] inorder, int inBegin, int inEnd){ 7 if(preBegin > preEnd) 8 return null; 9 int rootVal = preorder[preBegin]; 10 int i = inBegin; 11 for(; i <= inEnd; ++i){ 12 if(inorder[i] == rootVal) 13 break; 14 } 15 int leftLen = i - inBegin; 16 TreeNode root = new TreeNode(rootVal); 17 root.left = createTree(preorder, preBegin + 1, preBegin + leftLen ,inorder, inBegin, i - 1); //這里的邊界條件應該注意 18 root.right = createTree(preorder, preBegin + 1 + leftLen, preEnd, inorder, i + 1, inEnd); 19 return root; 20 } 21 }

?

轉載于:https://www.cnblogs.com/-wang-cheng/p/4910264.html

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