[leetcode] Construct Binary Tree from Inorder and Postorder Traversal

最新推荐文章于 2022-01-01 11:35:53 发布

Cabinathor

最新推荐文章于 2022-01-01 11:35:53 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/lmxmimihuhu/article/details/46508539

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本文介绍了一种利用中序和后序遍历序列重构二叉树的方法。通过递归方式确定根节点，并划分左右子树，最终构建完整的二叉树结构。

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From : https://leetcode.com/problems/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 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, postorder, 0, inorder.size()-1, 0, postorder.size()-1);
    }
    TreeNode* build(vector<int>& in, vector<int>& post, int si, int ei, int sp, int ep) {
        if(si>ei || sp>ep) return NULL;
        if(si==ei) return new TreeNode(in[si]);
        int rv=post[ep], inIdx=si;
        while(in[inIdx]!=rv) inIdx++;
        int postIdx=ep-ei+inIdx;
        TreeNode *root = new TreeNode(rv);
        root->left = build(in, post, si, inIdx-1, sp, postIdx-1);
        root->right = build(in, post, inIdx+1, ei, postIdx, ep-1);
        return root;
    }
};