题目

中序遍历和后序遍历树构造二叉树

根据中序遍历和后序遍历树构造二叉树

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

注意

你可以假设树中不存在相同数值的节点

解题

1.后序遍历最后一个结点就是根节点，根据这个根结点把中序遍历划分开来，同时也把后续遍历划分开来

2.递归就好了

程序感觉很简单不知道怎么写的，程序来源于九章

/**
* 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
*/
private int findPosition(int[] arr, int start, int end, int key) {
int i;
for (i = start; i <= end; i++) {
if (arr[i] == key) {
return i;
}
}
return -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, instart, inend, postorder\[postend\]);

    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;  
}

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);  
}  

}

Java Code

自己又实现了一遍，并加了注释

/**
* 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) {
// write your code here
if(inorder.length != postorder.length)
return null;
return buildTree(inorder,0,inorder.length -1 ,postorder,0,postorder.length -1 );
}
public int findroot(int[] inorder,int r){
for(int i=0;i iend)
return null;
int r = postorder[pend];
// 跟结点
TreeNode root = new TreeNode(r);
// 找到根节点
int l = findroot(inorder,r);
// 左子树 中序遍历 起始结束位置以此是：istart l-1
//后序遍历 起始位置是：pstart 结束位置:pstart(已经占据了一个位置所以要-1) + (左子树的长度) - 1
root.left = buildTree(inorder,istart,l-1,postorder,pstart,pstart+(l-1 - istart + 1) -1);
// 右子树 中序遍历 起始结束位置：l+1 iend
// 后序遍历 起始位置：pstart + (左子树的长度) ,结束位置 pend -1
root.right = buildTree(inorder,l+1,iend,postorder,pstart + (l-1-istart+1),pend -1);
return root;
}
}