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  • 106. Construct Binary Tree from Inorder and Postorder Traversal(js)

    106. Construct Binary Tree from Inorder and Postorder Traversal

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

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

    For example, given

    inorder = [9,3,15,20,7]
    postorder = [9,15,7,20,3]

    Return the following binary tree:

        3
       / 
      9  20
        /  
       15   7
    题意:通过中序遍历和后序遍历构建二叉搜索树
    代码如下:
    /**
     * Definition for a binary tree node.
     * function TreeNode(val) {
     *     this.val = val;
     *     this.left = this.right = null;
     * }
     */
    /**
     * @param {number[]} inorder
     * @param {number[]} postorder
     * @return {TreeNode}
     */
    var buildTree = function(inorder, postorder) {
                return backtrack(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
        }
    var backtrack = function( inorder,  inStart,inEnd , postorder, postStart, postEnd){
           if(inStart>inEnd || postStart>postEnd){
               return null;
           }
          let root = new TreeNode(postorder[postEnd]);
            let inIndex=0;
            for(let i=inStart;i<=inEnd;i++){
                if(inorder[i]==root.val){
                    inIndex=i;
                }
            }
            root.left=backtrack(inorder,inStart,inIndex-1,postorder,postStart,postStart+inIndex-inStart-1);
            root.right=backtrack(inorder,inIndex+1,inEnd,postorder,postStart+inIndex-inStart,postEnd-1);
            return root;
        }
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  • 原文地址:https://www.cnblogs.com/xingguozhiming/p/10713140.html
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