Construct Binary Tree from Inorder and Postorder Traversal
Desicription
Given inorder and postorder traversal of a tree, construct the binary tree.
Note: You may assume that duplicates do not exist in the tree.
Solution
代码语言：javascript复制/**
* 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 {
private:
TreeNode* generateTree(int postStart, int postEnd, int inStart, int inEnd, vector<int>& postorder, vector<int>& inorder) {
if(postStart > postEnd || inStart > inEnd)
return NULL;
TreeNode* root = new TreeNode(postorder[postEnd]);
int inIndex = 0;
for(int i = inStart; i <= inEnd; i ) {
if(inorder[i] == postorder[postEnd]){
inIndex = i;
break;
}
}
root->left = generateTree(postStart, postStart inIndex-inStart-1, inStart, inIndex-1, postorder, inorder);
root->right = generateTree(postEnd-inEnd inIndex, postEnd-1, inIndex 1, inEnd, postorder, inorder);
return root;
}
public:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
return generateTree(0, postorder.size()-1, 0, inorder.size()-1, postorder, inorder);
}
};
