leetcode102：二叉树的层序遍历 #47

Activity

1. sisterAn

   added

   LeetCode

   字节

   剑指Offer

   on May 21, 2020
2. plane-hjh commented on May 21, 2020

   

   plane-hjh

   on May 21, 2020

   Last edited by plane-hjh

   More actions

   这道题和二叉树的层次遍历相似，只需要把 reverse() 去除就可以了

   广度优先遍历

   var levelOrder = function(root) {
       if(!root) return []
       let res = [], 
           queue = [root]
       while(queue.length) {
           let curr = [],
               temp = []
           while(queue.length) {
               let node = queue.shift()
               curr.push(node.val)
               if(node.left) temp.push(node.left)
               if(node.right) temp.push(node.right)
           }
           res.push(curr)
           queue = temp
       }
       return res
   };

   深度优先遍历

   var levelOrder = function(root) {
       const res = []
       var dep = function (node, depth){
           if(!node) return
           res[depth] = res[depth]||[]
           res[depth].push(node.val)
           dep(node.left, depth + 1)
           dep(node.right, depth + 1)
       }
       dep(root, 0)
       return res
   };
3. AnranS commented on Nov 19, 2020

   

   AnranS

   on Nov 19, 2020

   More actions

   var levelOrder = function(root) {
       let res = [], queue = [];
       if(!root) return res;
       queue.push(root);
   
       while(queue.length) {
           let len = queue.length;
           res.push([]);
           for(let i = 0;i<len;i++) {
               let node = queue.shift();
               res[res.length - 1].push(node.val);
               if(node.left) queue.push(node.left);
               if(node.right) queue.push(node.right);
           }
       }
       return res;
   };
4. lewisYe commented on Apr 9, 2021

   

   lewisYe

   on Apr 9, 2021

   More actions

   使用广度优先遍历 然后再对数据进行分层

   var levelOrder = function(root) {
       if(!root || root.val == null) return []
      let queue = []
      let res = []
      queue.push(root)
      while(queue.length){
          let len = queue.length
          let curr = []
          for(var i =0;i<len;i++){
              let node = queue.shift()
              curr.push(node.val)
               if(node.left)queue.push(node.left)
               if(node.right)queue.push(node.right)
          }
          res.push(curr)
      }
       return res
   };