<LeetCode OJ> 105. Construct Binary Tree from Preorder and Inorder Traversal
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Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
分析:
首先要弄清楚什么是前序遍历和中序遍历?
前序遍历就是先遍历当前的节点,然后遍历该节点的左分支(即遍历左分支的所有节点),最后再遍历该节点的右分支。
中序遍历是先遍历当前节点的左分支,再遍历当前节点,最后遍历当前节点的右分支。
还有后序遍历,就是先左分支,再右分支,最后当前节点。
以下面的二叉树为例。
4
/ \
2 7
/ \ / \
1 3 6 9
其先序遍历结果:【4 2 1 3 7 6 9】
中序遍历结果是:【1 2 3 4 6 7 9】
对先序遍历来说:
先序遍历的每个值表示的结点都是接下来的若干个结点(根据中序遍历即确定)的父结点。
比如【4】是这个二叉树的根结点,【2】是【1 3】的父结点,【1】是 空的父结点,也即使叶子结点。
对中序遍历来说:
根结点(根据前序遍历即确定)一定在中间位置,该位置左边是其左子树,右边是其右子树。
比如【4】左边【1 2 3】全是根结点的左子树,右边是【6 7 9】是根结点的右子树。
对于【2】来说,【1】是其左子树,【3】是其右子树。
……
依次类推。
很明显,这是一个递归过程。
/** * 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>& preorder, vector<int>& inorder) { if(preorder.size()==0 || preorder.size()!=inorder.size()) return NULL; return helpBuild(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1); } TreeNode* helpBuild( vector<int> &preorder, int beginpre, int endpre, vector<int> &inorder, int beginord, int endord) { if(beginpre > endpre) return NULL; else if(beginpre == endpre) return new TreeNode(preorder[beginpre]); TreeNode* root = new TreeNode(preorder[beginpre]); int i = beginord; for(; i <= endord; i ++) { if(inorder[i] == preorder[beginpre]) break; } //inorder[i]此时就是根 int leftlen = i-beginord; //preorder[beginpre]是根 root->left = helpBuild(preorder, beginpre+1, beginpre+leftlen, inorder, beginord, beginord+leftlen-1); root->right = helpBuild(preorder, beginpre+leftlen+1, endpre, inorder, beginord+leftlen+1, endord); return root; }};
原文地址:http://blog.csdn.net/ebowtang/article/details/51580867
原作者博客:http://blog.csdn.net/ebowtang
本博客LeetCode题解索引:http://blog.csdn.net/ebowtang/article/details/50668895
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