LeetCode Construct Binary Tree from Inorder and Postorder Traversal
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题目
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
根据中序遍历和后续遍历构建树。
后续遍历某子树的最后一个元素是该子树的根,
在中序遍历中寻找相应元素,之前的元素在左子树上,之后的在右子树上。
根据数量对后序遍历序列进行划分,根据根的位置对中序遍历序列进行划分,递归。
代码:
/** * 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 *innerBuild(vector<int> &inorder,int ibegin,int iend,vector<int> &postorder,int pbegin,int pend){if(pbegin<pend){TreeNode *root=new TreeNode(postorder[pend-1]);//后续遍历的最后一个元素是根int i;for(i=ibegin;i<iend;i++)//中序遍历的相应元素前后分别为左右子树if(inorder[i]==postorder[pend-1])break;root->left=innerBuild(inorder,ibegin,i,postorder,pbegin,pbegin+i-ibegin);//递归root->right=innerBuild(inorder,i+1,iend,postorder,pbegin+i-ibegin,pend-1);return root;}return NULL;} TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { return innerBuild(inorder,0,inorder.size(),postorder,0,postorder.size()); }};
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