106.Construct Binary Tree from Inorder and Postorder Traversal

思路
  • 递归
/**
 * 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 {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return build(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);
    }
    TreeNode* build(vector<int>& inorder,int s0,int e0,vector<int>& postorder,int s1,int e1){
        if(s0 > e0 || s1 > e1) return NULL;
        int rootval = postorder[e1];
        TreeNode *root = new TreeNode(rootval);
        int mid = 0;
        for(int i = s0; i <= e0; i++){
            if(inorder[i] == rootval) mid = i;
        }
        root->left = build(inorder,s0,mid-1,postorder,s1,s1+mid-s0-1);
        root->right = build(inorder,mid+1,e0,postorder,s1+mid-s0,e1-1);
        return root;
    }
};
原文地址:https://www.cnblogs.com/UniMilky/p/7016254.html