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

Note:

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

the idea is similar with the former one [Leetcode-21]

java

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

c++

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

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