按照前序遍历和中序遍历构建二叉树

转载自：http://blog.csdn.net/sbitswc/article/details/26433051

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

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

There is an example.

        _______7______/              \__10__          ___2/      \        /4       3      _8\    /1  11

The preorder and inorder traversals for the binary tree above is:

preorder = {7,10,4,3,1,2,8,11}
inorder = {4,10,3,1,7,11,8,2}

The first node in preorder alwasy the root of the tree. We can break the tree like:
1st round:
preorder: {7}, {10,4,3,1}, {2,8,11}
inorder: {4,10,3,1}, {7}, {11, 8,2}

        _______7______/              \{4,10,3,1}       {11,8,2}

Since we alreay find that {7} will be the root, and in "inorder" sert, all the data in the left of {7} will construct the left sub-tree. And the right part will construct a right sub-tree. We can the left and right part agin based on the preorder.

2nd round
left part right part
preorder: {10}, {4}, {3,1} {2}, {8,11}
inorder: {4}, {10}, {3,1} {11,8}, {2}

        _______7______/              \__10__          ___2/      \        /4      {3,1}   {11,8}

see that, {10} will be the root of left-sub-tree and {2} will be the root of right-sub-tree.

Same way to split {3,1} and {11,8}, yo will get the complete tree now.

        _______7______/              \__10__          ___2/      \        /4       3      _8\    /1  11

So, simulate this process from bottom to top with recursion as following code.

c++

[cpp] view plain copy

TreeNode *BuildTreePI(
vector<int> &preorder,
vector<int> &inorder,
int p_s, int p_e,
int i_s, int i_e){
if(p_s > p_e) return NULL;
int pivot = preorder[p_s];
int i = i_s;
for(;i<i_e;i++){
if(inorder[i] == pivot)
break;
}
int length1 = i-i_s-1;
int length2 = i_e-i-1;
TreeNode* node = new TreeNode(pivot);
node->left = BuildTreePI(preorder,inorder,p_s+1,length1+p_s+1,i_s, i-1);
node->right = BuildTreePI(preorder, inorder, p_e-length2, p_e, i+1, i_e);
return node;
}
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
return BuildTreePI(preorder,inorder,0,preorder.size()-1,0,inorder.size()-1);
}

java

[java] view plain copy

public TreeNode buildTree(int[] preorder, int[] inorder) {
return buildPI(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1);
}
public TreeNode buildPI(int[] preorder, int[] inorder, int p_s, int p_e, int i_s, int i_e){
if(p_s>p_e)
return null;
int pivot = preorder[p_s];
int i = i_s;
for(;i<i_e;i++){
if(inorder[i]==pivot)
break;
}
TreeNode node = new TreeNode(pivot);
int lenLeft = i-i_s;
node.left = buildPI(preorder, inorder, p_s+1, p_s+lenLeft, i_s, i-1);
node.right = buildPI(preorder, inorder, p_s+lenLeft+1, p_e, i+1, i_e);
return node;
}

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