根据中序遍历和后序遍历树构造二叉树
注意事项
你可以假设树中不存在相同数值的节点
样例 给出树的中序遍历: [1,2,3] 和后序遍历: [1,3,2]
返回如下的树:
2
/ \
1 3
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
*@param inorder : A list of integers that inorder traversal of a tree
*@param postorder : A list of integers that postorder traversal of a tree
*@return : Root of a tree
*/
public TreeNode buildTree(int[] inorder, int[] postorder) {
if (inorder.length != postorder.length) {
return null;
}
return myBuildTree(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
}
private TreeNode myBuildTree(int[] inorder, int instart, int inend, int[] postorder, int poststart, int postend) {
if(instart > inend)
return null;
TreeNode root = new TreeNode(postorder[postend]);
int position = findposition(inorder, root.val);
root.left = myBuildTree(inorder, instart, position-1,postorder,poststart,poststart+position-instart-1);
root.right = myBuildTree(inorder, position+1,inend, postorder, poststart+position-instart,postend-1);
return root;
}
private int findposition(int[] inorder, int key) {
for(int i=0;i<inorder.length;i++) {
if(inorder[i] == key)
return i;
}
return -1;
}
}