leetcode 106 Construct Binary Tree from Inorder and Postorder Traversal

最新推荐文章于 2022-01-01 11:35:53 发布

walter1990

最新推荐文章于 2022-01-01 11:35:53 发布

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本文介绍了一种通过中序和后序遍历构建二叉树的方法。该方法利用递归过程，通过比较中序和后序遍历序列来构建树结构。文章提供了完整的C++实现代码。

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Given inorder and postorder 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> &postorder, int &num, vector<int> &inorder, int left, int right) {
		if(left > right) {
			root = NULL;
			return;
		}
		
		root = new TreeNode(postorder[num--]);

		int i = 0;

		for(i = left; i <= right; i++) {
			if(inorder[i]==postorder[num+1]) break;
		}

		tree(root->right, postorder, num, inorder, i+1, right);
		tree(root->left, postorder, num, inorder, left, i-1);

		return ;
	}
	TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
		TreeNode *root=NULL;
		if(inorder.size()==0 || postorder.size()==0) return root;

		int num = postorder.size()-1;

		tree(root, postorder, num, inorder, 0, num);
		return root;
	}
};