LeetCode——Construct Binary Tree from Inorder and Postorder Traversal

Question

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

参考：http://www.cnblogs.com/zhonghuasong/p/7096300.html

Code

/**

 * 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) {

        if (inorder.size() == 0 || postorder.size() == 0)

            return NULL;

        return ConstructTree(inorder, postorder, 0, inorder.size() - 1, 0, postorder.size() - 1);

    }

    TreeNode* ConstructTree(vector<int>& inorder, vector<int>& postorder,

            int in_start, int in_end, int post_start, int post_end) {

        int rootValue = postorder[post_end];

        TreeNode* root = new TreeNode(rootValue);

        if (in_start == in_end) {

            if (post_start == post_end && inorder[in_start] == postorder[post_start])

                return root;

        }

        int rootIn = in_start;

        while (rootIn <= in_end && inorder[rootIn] != rootValue)

            rootIn++;

        int leftPostLength = rootIn - in_start;

        if (leftPostLength > 0) {

            // 注意起点和终点的写法

            root->left = ConstructTree(inorder, postorder, in_start, rootIn - 1, post_start, post_start + leftPostLength - 1);

        }

        if (leftPostLength < in_end - in_start) {

            root->right = ConstructTree(inorder, postorder, rootIn + 1, in_end, post_start + leftPostLength, post_end - 1);

        }

        return root;

    }

};

巴特西

LeetCode——Construct Binary Tree from Inorder and Postorder Traversal

Question

Solution

Code

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