题目

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

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

题解

这道题跟pre+in一样的方法做,只不过找左子树右子树的位置不同而已。

         / \   

       / \ / \   

对于上图的树来说,
index: 0 1 2 3 4 5 6
中序遍历为: 4 2 5 1 6 3 7
后续遍历为: 4 5 2 6 7 3
为了清晰表示,我给节点上了颜色,红色是根节点,蓝色为左子树,绿色为右子树。
可以发现的规律是:
1. 中序遍历中根节点是左子树右子树的分割点。

2. 后续遍历的最后一个节点为根节点。 同样根据中序遍历找到根节点的位置,然后顺势计算出左子树串的长度。在后序遍历中分割出左子树串和右子树串,递归的建立左子树和右子树。
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return buildTree(inorder, 0, inorder.length-1, postorder, 0, postorder.length-1);
    }
    
        public TreeNode buildTree(int[] in, int inStart, int inEnd, int[] post, int postStart, int postEnd){
         if(inStart > inEnd || postStart > postEnd){
             return null;
        }
        int rootVal = post[postEnd];
        int rootIndex = 0;
        for(int i = inStart; i <= inEnd; i++){
             if(in[i] == rootVal){
                 rootIndex = i;
                 break;
             }
         }
       
         int len = rootIndex - inStart;
         TreeNode root = new TreeNode(rootVal);
         root.left = buildTree(in, inStart, rootIndex-1, post, postStart, postStart+len-1);
         root.right = buildTree(in, rootIndex+1, inEnd, post, postStart+len, postEnd-1);
       
        return root;
     }
												

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