LeetCode Construct Binary Tree from Preorder and Inorder Traversal

最新推荐文章于 2024-01-17 14:50:43 发布

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本文介绍了一种通过给定的先序和中序遍历序列来构建二叉树的方法。利用递归的方式找到根节点，并以此划分左右子树，最终完成整个二叉树的构建。

题目：

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

Note:
You may assume that duplicates do not exist in the tree.

分析参考：http://leetcode.com/2011/04/construct-binary-tree-from-inorder-and-preorder-postorder-traversal.html

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        TreeNode *root = createTree(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
		return root;
    }
private:
	TreeNode *createTree(vector<int> &preorder, int preBeg, int preEnd, vector<int> &inorder, int inBeg, int inEnd) {
		if(preBeg > preEnd)
			return NULL;
		int rootVal = preorder[preBeg];
		TreeNode *root = new TreeNode(rootVal);
		//找到rootVal在中序中的索引 
		int index;
		for(int i = inBeg; i <= inEnd; i++) {
			if(inorder[i] == rootVal) {
				index = i;
				break;	
			}	
		}
		int length = index - inBeg;
		root->left = createTree(preorder, preBeg+1, preBeg+length, inorder, inBeg, index-1);
		root->right = createTree(preorder, preBeg+length+1, preEnd, inorder, index+1, inEnd);
		return root;
	}    
};