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.

 

/**
 * 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 *buildTree(vector<int> &inorder, vector<int> &postorder) {
        if(inorder.size() != postorder.size()) return NULL;
        int n = inorder.size();
        return build(inorder, 0, postorder, n-1, n);
    }
    
    TreeNode *build(vector<int> &io, int inStart, vector<int> &po, int postEnd, int len) {
        if(inStart < 0 || inStart + len > io.size() || postEnd >= po.size() || postEnd - len < -1 || len < 1) {
            return NULL;
        }
        
        int rootVal = po[postEnd];
        
        int i = inStart;
        while(io[i] != rootVal) ++i;
        
        TreeNode *root = new TreeNode(rootVal);
        
        int leftLen = i - inStart;
        int rightLen = len - leftLen - 1;
        
        root->right = build(io, i+1, po, postEnd - 1, rightLen);
        root->left = build(io, inStart, po, postEnd - rightLen - 1, leftLen);
        
        return root;
    }
};

 

 

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