[LeetCode] Construct Binary Tree from Inorder and Postorder Traversal

eternalwaiting

于 2014-02-26 11:57:08 发布

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文章标签： leetcode binary tree recursion

本文链接：https://blog.youkuaiyun.com/eternalwaiting/article/details/19971971

本文介绍了一种通过给定的二叉树中序遍历和后序遍历序列来重建该二叉树的方法。核心思路在于利用后序遍历的最后一个元素作为根节点，并在中序遍历中找到该根节点的位置，以此划分左右子树。随后递归地对左右子树应用相同的操作。

1. The last element in postorder is the root;

2. Find the root in inorder, the left side belongs to left subtree, the right side belongs to right subtree;

3. Repeat step 1 and 2 on the left side elements and the right side elements recursively.

/**
 * 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 *buildHelper(vector<int> &inorder, vector<int> &postorder, int ib,int ie, int pb, int pe)
    {
        if(ib>ie || pb>pe) return NULL;
        TreeNode *root = new TreeNode(postorder[pe]);

        int i;
        for(i=ib;i<=ie;i++)
            if(inorder[i]==postorder[pe])
                break;
        root->left = buildHelper(inorder,postorder,ib,i-1,pb,pb+i-ib-1);
        root->right = buildHelper(inorder,postorder,i+1,ie,pb+i-ib,pe-1);
        return root;
    }
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        return buildHelper(inorder,postorder,0,inorder.size()-1,0,postorder.size()-1);
    }
};