树Populating Next Right Pointers in Each Node
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题目:
Given a binary tree
struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; }
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL
.
Initially, all next pointers are set to NULL
.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1 / 2 3 / \ / 4 5 6 7
After calling your function, the tree should look like:
1 -> NULL / 2 -> 3 -> NULL / \ / 4->5->6->7 -> NULL
思路:
层序遍历二叉树,把每一层中前一个节点的next指向后一个节点,使用队列辅助层序遍历时,在队列中用NULL来分割每层的节点。
/** * Definition for binary tree with next pointer. * function TreeLinkNode(val) { * this.val = val; * this.left = this.right = this.next = null; * } */ /** * @param {TreeLinkNode} root * @return {void} Do not return anything, modify tree in-place instead. */ var connect = function(root) { if(root==null){ return; } var stack=[],pre=null; stack.push(root); stack.push(null); while(stack.length!=0){ var p=stack.unshift(); if(p!=null){ if(p.left){ stack.push(p.left); } if(p.right){ stack.push(p.right); } }else{ if(stack.length!=0){ stack.push(null); } } if(pre!=null){ pre.next=p; } pre=p; } };
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