首页 > 代码库 > leetcode 刷题之路 64 Construct Binary Tree from Inorder and Postorder Traversal

leetcode 刷题之路 64 Construct Binary Tree from Inorder and Postorder Traversal

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

Note:

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

给出二叉树的中序遍历和后序遍历结果,恢复出二叉树。

后序遍历序列的最后一个元素值是二叉树的根节点的值,查找该元素在中序遍历序列中的位置mid,根据中序遍历和后序遍历性质,有:

位置mid以前的序列部分为二叉树根节点左子树中序遍历的结果,得出该序列的长度n,则后序遍历序列前n个元素为二叉树根节点左子树后序遍历的结果,由这两个中序遍历和后序遍历子序列恢复出左子树;

位置mid以后的序列部分为二叉树根节点右子树中序遍历的结果,得出该序列的长度m,则后序遍历序列(除去最后一个元素)后m个元素为二叉树根节点右子树后序遍历的结果,由这两个中序遍历和后序遍历子序列恢复出左子树;

以上描述中递归地引用了由中序遍历和后序遍历恢复子树的部分,因此程序也采用递归实现。

AC code:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 

 TreeNode *helper(vector<int> &inorder, int b1, int e1, vector<int> &postorder, int b2, int e2)
 {
	 if (b1>e1)
		 return NULL;
	 int mid;
	 for (int i = b1; i <= e1; i++)
		 if (inorder[i] == postorder[e2])
		 {
			 mid = i;
			 break;
		 }
			 

	 TreeNode* root = new TreeNode(inorder[mid]);
	 root->left = helper(inorder, b1, mid - 1, postorder, b2, b2 + mid - b1-1);
	 root->right = helper(inorder, mid + 1, e1, postorder, b2 + mid-b1, e2 - 1);
	 return root;
 }
 class Solution {
 public:
	 TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder)
	 {
		 return helper(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);
	 }
 };