leetcode 105 Constract Binary Tree from Preorder and Inorder Traversal

walter1990

于 2016-06-05 11:07:16 发布

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本文介绍了一种通过前序遍历和中序遍历构建二叉树的方法。利用递归技术，该方法能有效地构建出不含重复节点的二叉树。

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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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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void tree(TreeNode *&root, vector<int> &preorder, int &num, vector<int> &inorder, int left, int right) {
		if(left > right) {
			root = NULL;
			return ;
		}
		root = new TreeNode(preorder[num++]);
		int temp = preorder[num-1], i;

		for(i=left; i <= right; i++) {
			if(inorder[i]==temp) break;
		}
		tree(root->left, preorder, num, inorder, left, i-1);
		tree(root->right, preorder, num, inorder, i+1, right);
	}
	TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
		TreeNode *root=NULL;
		if(preorder.size()==0 || inorder.size()==0) return root;
		
		int num = 0, i=0, left=0, right=inorder.size();
		tree(root, preorder, num, inorder, left, right-1);
		return root;
	}
};