我们都知道，已知中序和后序的序列是可以唯一确定一个二叉树的。
初始化时候二叉树为：==================
中序遍历序列， ======O===========
后序遍历序列， =================O
红色部分是左子树，黑色部分是右子树，O是根节点
如上图所示，O是根节点，由后序遍历可知，
根据这个O可以把找到其在中序遍历当中的位置，进而，知道当前这个根节点O的左子树的前序遍历和中序遍历序列的范围。
以及右子树的前序遍历和中序遍历序列的范围。
到这里返现出现了重复的子问题，而且子问题的规模没有原先的问题大，即红色部分和黑色部分。
而联系这两个子问题和原先的大问题的纽带是这个找到的根节点。
可以选择用递归来解决这个问题，递归的结束条件是子问题序列里面只有一个元素。
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给定一个二叉树的中序和后序遍历序列，构造这个二叉树。
笔记：
你可以假定，这棵树里面没有重复的节点。
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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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test.cpp：
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#include <iostream>
#include <cstdio> #include <stack> #include <vector> #include "BinaryTree.h" using namespace std; /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ TreeNode *build(vector<int> &inorder, int left1, int right1, vector<int> &postorder, int left2, int right2) { if(right1 - left1 != right2 - left2) { return NULL; } if(right1 >= inorder.size() || right2 >= postorder.size()) { return NULL; } //递归结束条件 if(left1 == right1 && left2 == right2) { TreeNode *root = new TreeNode(inorder[left1]); return root; } else if(left1 < right1 && left2 < right2) { TreeNode *root = new TreeNode(postorder[right2]); int i; for(i = right1; i >= left1; i--) { //找到中序遍历的根节点 if(inorder[i] == postorder[right2]) { break; } } if(i < left1) { return NULL; } root->left = build(inorder, left1, i - 1, postorder, left2, right2 + i - right1 - 1); root->right = build(inorder, i + 1, right1, postorder, right2 + i - right1, right2 - 1); return root; } else { return NULL; } } TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { return build(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1); } vector<vector<int> > levelOrder(TreeNode *root) { vector<vector<int> > matrix; if(root == NULL) { return matrix; } vector<int> temp; temp.push_back(root->val); matrix.push_back(temp); vector<TreeNode *> path; path.push_back(root); int count = 1; while(!path.empty()) { TreeNode *tn = path.front(); if(tn->left) { path.push_back(tn->left); } if(tn->right) { path.push_back(tn->right); } path.erase(path.begin()); count--; if(count == 0) { vector<int> tmp; vector<TreeNode *>::iterator it = path.begin(); for(; it != path.end(); ++it) { tmp.push_back((*it)->val); } if(tmp.size() > 0) { matrix.push_back(tmp); } count = path.size(); } } return matrix; } // 树中结点含有分叉， // 6 // / // 7 2 // / // 1 4 // / // 3 5 int main() { TreeNode *pNodeA1 = CreateBinaryTreeNode(6); TreeNode *pNodeA2 = CreateBinaryTreeNode(7); TreeNode *pNodeA3 = CreateBinaryTreeNode(2); TreeNode *pNodeA4 = CreateBinaryTreeNode(1); TreeNode *pNodeA5 = CreateBinaryTreeNode(4); TreeNode *pNodeA6 = CreateBinaryTreeNode(3); TreeNode *pNodeA7 = CreateBinaryTreeNode(5); ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3); ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5); ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7); int in[7] = {1, 7, 3, 4, 5, 6, 2}; int post[7] = {1, 3, 5, 4, 7, 2, 6}; vector<int> inorder(in, in + 7), postorder(post, post + 7); TreeNode *root = buildTree(inorder, postorder); vector<vector<int> > ans = levelOrder(root); for (int i = 0; i < ans.size(); ++i) { for (int j = 0; j < ans[i].size(); ++j) { cout << ans[i][j] << " "; } cout << endl; } DestroyTree(root); return 0; } |
结果输出：
6
7 2
1 4
3 5
ps.测试的输出用的是层次遍历
BinaryTree.h:
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#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_ struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; TreeNode *CreateBinaryTreeNode(int value); void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight); void PrintTreeNode(TreeNode *pNode); void PrintTree(TreeNode *pRoot); void DestroyTree(TreeNode *pRoot); #endif /*_BINARY_TREE_H_*/ |
BinaryTree.cpp:
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#include <iostream>
#include <cstdio> #include "BinaryTree.h" using namespace std; /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ //创建结点 TreeNode *CreateBinaryTreeNode(int value) { TreeNode *pNode = new TreeNode(value); return pNode; } //连接结点 void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight) { if(pParent != NULL) { pParent->left = pLeft; pParent->right = pRight; } } //打印节点内容以及左右子结点内容 void PrintTreeNode(TreeNode *pNode) { if(pNode != NULL) { printf("value of this node is: %d ", pNode->val); if(pNode->left != NULL) printf("value of its left child is: %d. ", pNode->left->val); else printf("left child is null. "); if(pNode->right != NULL) printf("value of its right child is: %d. ", pNode->right->val); else printf("right child is null. "); } else { printf("this node is null. "); } printf(" "); } //前序遍历递归方法打印结点内容 void PrintTree(TreeNode *pRoot) { PrintTreeNode(pRoot); if(pRoot != NULL) { if(pRoot->left != NULL) PrintTree(pRoot->left); if(pRoot->right != NULL) PrintTree(pRoot->right); } } void DestroyTree(TreeNode *pRoot) { if(pRoot != NULL) { TreeNode *pLeft = pRoot->left; TreeNode *pRight = pRoot->right; delete pRoot; pRoot = NULL; DestroyTree(pLeft); DestroyTree(pRight); } } |