【leetcode】Construct Binary Tree from Inorder and Postorder Traversal

最新推荐文章于 2021-02-03 15:56:42 发布

原创最新推荐文章于 2021-02-03 15:56:42 发布 · 446 阅读

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#leetcode #算法 #二叉树 #遍历 #栈

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本文详细介绍了如何通过中根遍历和后根遍历来构造二叉树，包括算法思路和具体实现步骤。

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

Note:

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

Accept: 22ms

思路：

中根遍历+后根遍历还原一棵二叉树。
top为当前栈顶元素，然后从中根里找到对应的元素的位置，作为新的中根元素，与top里记录的当前中根左右两端做比较：
- 如果新的中根在右边，则插入到右子树，并且新中根入栈；
- 如果新的中根在左边，则插入到左子树，并且新中根入栈；
- 否则，弹出top，然后重新比较左右2端。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

struct Elem {
  struct TreeNode* node;
  int left_beg;
  int left_end;
  int right_beg;
  int right_end;
};

int find(int* inorder, int left, int right, int val) {
  for (int i = left; i < right; ++i) {
    if (inorder[i] == val) {
      return i;
    }
  }
  return -1;
}

int push(struct Elem* stack, int* p, struct TreeNode *node,
         int left, int mid, int right) {
  stack[*p].node = node;
  stack[*p].left_beg = left;
  stack[*p].left_end = mid;
  stack[*p].right_beg = mid + 1;
  stack[*p].right_end = right;
  ++*p;
}

struct TreeNode* makeNode(int val) {
  struct TreeNode* node = malloc(sizeof(struct TreeNode));
  node->val = val;
  node->left = NULL;
  node->right = NULL;
  return node;
}

struct TreeNode *buildTree(int *inorder, int *postorder, int n) {
  if (n == 0) {
    return NULL;
  }

  const int max_size = 10000;
  struct Elem stack[max_size];
  int p = 0;

  struct TreeNode* head = makeNode(postorder[n-1]);
  int mid = find(inorder, 0, n, head->val);
  push(stack, &p, head, 0, mid, n);

  int i = n - 2;
  while (i >= 0) {
    int val = postorder[i];
    struct Elem *top = &stack[p - 1];

    int mid = find(inorder, top->right_beg, top->right_end, val);
    if (mid >= 0) {
      struct TreeNode *node = makeNode(val);
      top->node->right = node;
      push(stack, &p, node, top->right_beg, mid, top->right_end);
      --i;
      continue;
    }

    mid = find(inorder, top->left_beg, top->left_end, val);
    if (mid >= 0) {
      struct TreeNode *node = makeNode(val);
      top->node->left = node;
      push(stack, &p, node, top->left_beg, mid, top->left_end);
      --i;
      continue;
    }

    // pop:
    p--;
  } // while

  return head;
}