Construct Binary Tree from Inorder and Postorder Traversal

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Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

#include<stdio.h>
#include<stdlib.h>

typedef struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
}TreeNode;

TreeNode *buildTreeCore(int *inorder, int inbegin, int inend, int *postorder, int postbegin, int postend) {
    TreeNode *root = (TreeNode *)malloc(sizeof(TreeNode));
    root->val = postorder[postend];
    root->left = root->right = NULL;
    if(inbegin == inend || postbegin == postend) return root;
    int i;
    for(i = inend; i >= inbegin && postorder[postend] != inorder[i]; i--) ;
    if(i < inend) {
        root->right = buildTreeCore(inorder, i + 1, inend, postorder, postbegin + i - inbegin, postend - 1);
    }
    if(i > inbegin) {
        root->left = buildTreeCore(inorder, inbegin, i - 1, postorder, postbegin, postbegin + i - inbegin - 1);
    }
    return root;
}

TreeNode *buildTree(int *inorder, int *postorder, int n) {
    if(inorder == NULL || postorder == NULL) return NULL;
    TreeNode *root = buildTreeCore(inorder, 0, n - 1, postorder, 0, n - 1);
}

void Travel(TreeNode *root) {
    if(root != NULL) {
        printf("%d, ", root->val);
        Travel(root->left);
        Travel(root->right);
    }
}

void main() {
    int in[] = {1, 2, 3, 4, 5, 6};
    int post[] = {1, 2, 4, 5, 6, 3};
    TreeNode *root = buildTree(in, post, 6);
    Travel(root);
    printf("\n");
}

Construct Binary Tree from Inorder and Postorder Traversal

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原文地址：http://blog.csdn.net/uj_mosquito/article/details/43309807